Check convergence of the series $sum_n=1^inftycos^n^3frac1sqrt n$ [closed]

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up vote
3
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favorite

Prove the convergence of the series
$$sum_n=1^inftycos^n^3frac1sqrt n$$
This is the first time that I'm learning about the convergence of the series and there are so many theorems about how to prove one and I really don't know which one to use.

I would really appreciate some help.

sequences-and-series convergence

share|cite|improve this question

edited Jul 25 at 13:55

asked Jul 25 at 13:49

J.Dane

159112

closed as off-topic by Did, Nosrati, amWhy, Xander Henderson, Parcly Taxel Jul 26 at 1:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Nosrati, amWhy, Xander Henderson, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does $n3$ mean? It makes a huge difference whether you mean $3n$ or $n^3$.
  – B. Goddard
  Jul 25 at 13:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it's $n^3$ . I edited it
  – J.Dane
  Jul 25 at 13:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don't think there are "so many" theorems. When you have an $n$th power (or worse) then the root test suggests itself.
  – B. Goddard
  Jul 25 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I said I'm new to this. I am trying to learn which theorem should I use for specific problems
  – J.Dane
  Jul 25 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @J.Dane When proving convergence of series, one of the best tools in my opinion is using the comparison test. There are a bunch of series (geometric, harmonic, alternating harmonic, etc) which have already been proven to converge or diverge. If you are looking to prove convergence for a series A, you can compare it to a series B where every term in B is larger than the the terms in A. If B converges, then A must also converge. Hope this helps!
  – user9750060
  Jul 25 at 14:15
  

  
  
  
  

 | 
show 2 more comments

up vote
3
down vote

favorite

Prove the convergence of the series
$$sum_n=1^inftycos^n^3frac1sqrt n$$
This is the first time that I'm learning about the convergence of the series and there are so many theorems about how to prove one and I really don't know which one to use.

I would really appreciate some help.

sequences-and-series convergence

share|cite|improve this question

edited Jul 25 at 13:55

asked Jul 25 at 13:49

J.Dane

159112

closed as off-topic by Did, Nosrati, amWhy, Xander Henderson, Parcly Taxel Jul 26 at 1:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Nosrati, amWhy, Xander Henderson, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does $n3$ mean? It makes a huge difference whether you mean $3n$ or $n^3$.
  – B. Goddard
  Jul 25 at 13:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it's $n^3$ . I edited it
  – J.Dane
  Jul 25 at 13:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don't think there are "so many" theorems. When you have an $n$th power (or worse) then the root test suggests itself.
  – B. Goddard
  Jul 25 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I said I'm new to this. I am trying to learn which theorem should I use for specific problems
  – J.Dane
  Jul 25 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @J.Dane When proving convergence of series, one of the best tools in my opinion is using the comparison test. There are a bunch of series (geometric, harmonic, alternating harmonic, etc) which have already been proven to converge or diverge. If you are looking to prove convergence for a series A, you can compare it to a series B where every term in B is larger than the the terms in A. If B converges, then A must also converge. Hope this helps!
  – user9750060
  Jul 25 at 14:15
  

  
  
  
  

 | 
show 2 more comments

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Prove the convergence of the series
$$sum_n=1^inftycos^n^3frac1sqrt n$$
This is the first time that I'm learning about the convergence of the series and there are so many theorems about how to prove one and I really don't know which one to use.

I would really appreciate some help.

sequences-and-series convergence

share|cite|improve this question

edited Jul 25 at 13:55

asked Jul 25 at 13:49

J.Dane

159112

Prove the convergence of the series
$$sum_n=1^inftycos^n^3frac1sqrt n$$
This is the first time that I'm learning about the convergence of the series and there are so many theorems about how to prove one and I really don't know which one to use.

I would really appreciate some help.

sequences-and-series convergence

share|cite|improve this question

edited Jul 25 at 13:55

asked Jul 25 at 13:49

J.Dane

159112

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 13:55

edited Jul 25 at 13:55

edited Jul 25 at 13:55

asked Jul 25 at 13:49

J.Dane

159112

asked Jul 25 at 13:49

J.Dane

159112

asked Jul 25 at 13:49

J.Dane

159112

159112

closed as off-topic by Did, Nosrati, amWhy, Xander Henderson, Parcly Taxel Jul 26 at 1:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Nosrati, amWhy, Xander Henderson, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Did, Nosrati, amWhy, Xander Henderson, Parcly Taxel Jul 26 at 1:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Nosrati, amWhy, Xander Henderson, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does $n3$ mean? It makes a huge difference whether you mean $3n$ or $n^3$.
  – B. Goddard
  Jul 25 at 13:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it's $n^3$ . I edited it
  – J.Dane
  Jul 25 at 13:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don't think there are "so many" theorems. When you have an $n$th power (or worse) then the root test suggests itself.
  – B. Goddard
  Jul 25 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I said I'm new to this. I am trying to learn which theorem should I use for specific problems
  – J.Dane
  Jul 25 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @J.Dane When proving convergence of series, one of the best tools in my opinion is using the comparison test. There are a bunch of series (geometric, harmonic, alternating harmonic, etc) which have already been proven to converge or diverge. If you are looking to prove convergence for a series A, you can compare it to a series B where every term in B is larger than the the terms in A. If B converges, then A must also converge. Hope this helps!
  – user9750060
  Jul 25 at 14:15
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does $n3$ mean? It makes a huge difference whether you mean $3n$ or $n^3$.
  – B. Goddard
  Jul 25 at 13:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it's $n^3$ . I edited it
  – J.Dane
  Jul 25 at 13:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don't think there are "so many" theorems. When you have an $n$th power (or worse) then the root test suggests itself.
  – B. Goddard
  Jul 25 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I said I'm new to this. I am trying to learn which theorem should I use for specific problems
  – J.Dane
  Jul 25 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @J.Dane When proving convergence of series, one of the best tools in my opinion is using the comparison test. There are a bunch of series (geometric, harmonic, alternating harmonic, etc) which have already been proven to converge or diverge. If you are looking to prove convergence for a series A, you can compare it to a series B where every term in B is larger than the the terms in A. If B converges, then A must also converge. Hope this helps!
  – user9750060
  Jul 25 at 14:15
  

  
  
  
  

What does $n3$ mean? It makes a huge difference whether you mean $3n$ or $n^3$.
– B. Goddard
Jul 25 at 13:52

What does $n3$ mean? It makes a huge difference whether you mean $3n$ or $n^3$.
– B. Goddard
Jul 25 at 13:52

it's $n^3$ . I edited it
– J.Dane
Jul 25 at 13:56

it's $n^3$ . I edited it
– J.Dane
Jul 25 at 13:56

1

1

I don't think there are "so many" theorems. When you have an $n$th power (or worse) then the root test suggests itself.
– B. Goddard
Jul 25 at 13:57

I don't think there are "so many" theorems. When you have an $n$th power (or worse) then the root test suggests itself.
– B. Goddard
Jul 25 at 13:57

I said I'm new to this. I am trying to learn which theorem should I use for specific problems
– J.Dane
Jul 25 at 13:59

I said I'm new to this. I am trying to learn which theorem should I use for specific problems
– J.Dane
Jul 25 at 13:59

@J.Dane When proving convergence of series, one of the best tools in my opinion is using the comparison test. There are a bunch of series (geometric, harmonic, alternating harmonic, etc) which have already been proven to converge or diverge. If you are looking to prove convergence for a series A, you can compare it to a series B where every term in B is larger than the the terms in A. If B converges, then A must also converge. Hope this helps!
– user9750060
Jul 25 at 14:15

@J.Dane When proving convergence of series, one of the best tools in my opinion is using the comparison test. There are a bunch of series (geometric, harmonic, alternating harmonic, etc) which have already been proven to converge or diverge. If you are looking to prove convergence for a series A, you can compare it to a series B where every term in B is larger than the the terms in A. If B converges, then A must also converge. Hope this helps!
– user9750060
Jul 25 at 14:15

 | 
show 2 more comments

4 Answers
4

active

oldest

votes

up vote
1
down vote



accepted

We need neither the root test nor Taylor's Theorem to proceed. Here instead, we use the elementary inequalities $log(1-x)le -x$ and $sin(x)ge 2x/pi$ (for $0<xle pi/2$).

---

Proceeding, we see that for $nge1$

$$beginalign
0le cos^n^3(n^-1/2)&=e^n^3log(1-2sin^2(n^-1/2/2))\\
&le e^-2n^3sin^2(n^-1/2/2)\\
&le e^-2n^2/pi^2\\
endalign$$

Inasmuch as $sum_n=1^infty e^-2n^2/pi^2$ converges, we conclude that the series of interest converges also.

share|cite|improve this answer

edited Jul 25 at 14:58

answered Jul 25 at 14:48

Mark Viola

126k1172167

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very nice and elegant solution!
  – gimusi
  Jul 25 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Thank you! Much appreciated.
  – Mark Viola
  Jul 25 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would the down voter care to comment???'
  – Mark Viola
  Jul 26 at 21:11
  

  
  
  
  

add a comment | 

up vote
2
down vote

We have that

$$left(cosleft(frac1sqrt nright)right)^n^3= left(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)^n^3=e^n^3logleft(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)sim e^-fracn^22$$

therefore the given series converges by limit comparison test with $sum e^-fracn^22$.

As an alternative by root test for $a_n=left(cosleft(frac1sqrt nright)right)^n^3$ we have

$$sqrt[n]a_n=left(cosleft(frac1sqrt nright)right)^n^2sim e^-fracn2to 0$$

share|cite|improve this answer

edited Jul 25 at 14:16

answered Jul 25 at 13:59

gimusi

65k73583

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don’t understand the downvote.
  – Szeto
  Jul 25 at 14:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve?
  – gimusi
  Jul 25 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it got downvoted, because the solution is not elementary enough?
  – Cornman
  Jul 25 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note there are two close votes on the question (for a totally silly reason, imo). It can happen that an answer gets downvoted simply bbecause people disapprove of answering "bad" questions...
  – David C. Ullrich
  Jul 25 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) math.stackexchange.com/q/2862389/505767
  – gimusi
  Jul 25 at 14:29
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

The simplest here is to use the root test: you should find the
$$lim_ntoinftybiggl(cos^n^3!frac1sqrt nbiggr)^!tfrac1n=lim_ntoinftycos^n^2!frac1sqrt n=0.$$

Hint:

This is equivalent to showing $;lim_ntoinftyn^2logbiggl(cosdfrac1sqrt nbiggr)=-infty$, and you can use for that Taylor formula at order $2$:
$$cos u=1-fracu^22+o(u^2).$$

share|cite|improve this answer

edited Jul 25 at 14:36

answered Jul 25 at 14:24

Bernard

110k635103

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it should be $n^2$ not $1/n^2$
  – J.Dane
  Jul 25 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich: You're right. I should always check computations with a pencil and paper before posting, but I didn't. I've fixed the answer. Thank you for pointing the error!
  – Bernard
  Jul 25 at 14:39
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

In style to @MarkViola's version. Here is a proof of
$$0leqcosxleq e^-fracx^22, forall xin left[0,fracpi2right]$$
using nothing but derivatives, so may be classified as elementary.
Also, $0<frac1sqrtn<fracpi2, forall n >0$. Then
$$0leqcosfrac1sqrtnleq e^-frac12nRightarrow
0leq left(cosfrac1sqrtnright)^n^3leq e^-fracn^22 tag1$$
and finally
$$0leq sumlimits_n=1 left(cosfrac1sqrtnright)^n^3leq sumlimits_n=1e^-fracn^22$$

share|cite|improve this answer

answered Jul 25 at 15:28

rtybase

8,83721333

add a comment | 

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

We need neither the root test nor Taylor's Theorem to proceed. Here instead, we use the elementary inequalities $log(1-x)le -x$ and $sin(x)ge 2x/pi$ (for $0<xle pi/2$).

---

Proceeding, we see that for $nge1$

$$beginalign
0le cos^n^3(n^-1/2)&=e^n^3log(1-2sin^2(n^-1/2/2))\\
&le e^-2n^3sin^2(n^-1/2/2)\\
&le e^-2n^2/pi^2\\
endalign$$

Inasmuch as $sum_n=1^infty e^-2n^2/pi^2$ converges, we conclude that the series of interest converges also.

share|cite|improve this answer

edited Jul 25 at 14:58

answered Jul 25 at 14:48

Mark Viola

126k1172167

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very nice and elegant solution!
  – gimusi
  Jul 25 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Thank you! Much appreciated.
  – Mark Viola
  Jul 25 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would the down voter care to comment???'
  – Mark Viola
  Jul 26 at 21:11
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

We need neither the root test nor Taylor's Theorem to proceed. Here instead, we use the elementary inequalities $log(1-x)le -x$ and $sin(x)ge 2x/pi$ (for $0<xle pi/2$).

---

Proceeding, we see that for $nge1$

$$beginalign
0le cos^n^3(n^-1/2)&=e^n^3log(1-2sin^2(n^-1/2/2))\\
&le e^-2n^3sin^2(n^-1/2/2)\\
&le e^-2n^2/pi^2\\
endalign$$

Inasmuch as $sum_n=1^infty e^-2n^2/pi^2$ converges, we conclude that the series of interest converges also.

share|cite|improve this answer

edited Jul 25 at 14:58

answered Jul 25 at 14:48

Mark Viola

126k1172167

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very nice and elegant solution!
  – gimusi
  Jul 25 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Thank you! Much appreciated.
  – Mark Viola
  Jul 25 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would the down voter care to comment???'
  – Mark Viola
  Jul 26 at 21:11
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

We need neither the root test nor Taylor's Theorem to proceed. Here instead, we use the elementary inequalities $log(1-x)le -x$ and $sin(x)ge 2x/pi$ (for $0<xle pi/2$).

---

Proceeding, we see that for $nge1$

$$beginalign
0le cos^n^3(n^-1/2)&=e^n^3log(1-2sin^2(n^-1/2/2))\\
&le e^-2n^3sin^2(n^-1/2/2)\\
&le e^-2n^2/pi^2\\
endalign$$

Inasmuch as $sum_n=1^infty e^-2n^2/pi^2$ converges, we conclude that the series of interest converges also.

share|cite|improve this answer

edited Jul 25 at 14:58

answered Jul 25 at 14:48

Mark Viola

126k1172167

We need neither the root test nor Taylor's Theorem to proceed. Here instead, we use the elementary inequalities $log(1-x)le -x$ and $sin(x)ge 2x/pi$ (for $0<xle pi/2$).

---

Proceeding, we see that for $nge1$

$$beginalign
0le cos^n^3(n^-1/2)&=e^n^3log(1-2sin^2(n^-1/2/2))\\
&le e^-2n^3sin^2(n^-1/2/2)\\
&le e^-2n^2/pi^2\\
endalign$$

Inasmuch as $sum_n=1^infty e^-2n^2/pi^2$ converges, we conclude that the series of interest converges also.

share|cite|improve this answer

edited Jul 25 at 14:58

answered Jul 25 at 14:48

Mark Viola

126k1172167

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 14:58

edited Jul 25 at 14:58

edited Jul 25 at 14:58

answered Jul 25 at 14:48

Mark Viola

126k1172167

answered Jul 25 at 14:48

Mark Viola

126k1172167

answered Jul 25 at 14:48

Mark Viola

126k1172167

126k1172167

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very nice and elegant solution!
  – gimusi
  Jul 25 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Thank you! Much appreciated.
  – Mark Viola
  Jul 25 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would the down voter care to comment???'
  – Mark Viola
  Jul 26 at 21:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Very nice and elegant solution!
  – gimusi
  Jul 25 at 14:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Thank you! Much appreciated.
  – Mark Viola
  Jul 25 at 15:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would the down voter care to comment???'
  – Mark Viola
  Jul 26 at 21:11
  

  
  
  
  

Very nice and elegant solution!
– gimusi
Jul 25 at 14:59

Very nice and elegant solution!
– gimusi
Jul 25 at 14:59

@gimusi Thank you! Much appreciated.
– Mark Viola
Jul 25 at 15:00

@gimusi Thank you! Much appreciated.
– Mark Viola
Jul 25 at 15:00

Would the down voter care to comment???'
– Mark Viola
Jul 26 at 21:11

Would the down voter care to comment???'
– Mark Viola
Jul 26 at 21:11

add a comment | 

up vote
2
down vote

We have that

$$left(cosleft(frac1sqrt nright)right)^n^3= left(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)^n^3=e^n^3logleft(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)sim e^-fracn^22$$

therefore the given series converges by limit comparison test with $sum e^-fracn^22$.

As an alternative by root test for $a_n=left(cosleft(frac1sqrt nright)right)^n^3$ we have

$$sqrt[n]a_n=left(cosleft(frac1sqrt nright)right)^n^2sim e^-fracn2to 0$$

share|cite|improve this answer

edited Jul 25 at 14:16

answered Jul 25 at 13:59

gimusi

65k73583

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don’t understand the downvote.
  – Szeto
  Jul 25 at 14:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve?
  – gimusi
  Jul 25 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it got downvoted, because the solution is not elementary enough?
  – Cornman
  Jul 25 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note there are two close votes on the question (for a totally silly reason, imo). It can happen that an answer gets downvoted simply bbecause people disapprove of answering "bad" questions...
  – David C. Ullrich
  Jul 25 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) math.stackexchange.com/q/2862389/505767
  – gimusi
  Jul 25 at 14:29
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

We have that

$$left(cosleft(frac1sqrt nright)right)^n^3= left(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)^n^3=e^n^3logleft(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)sim e^-fracn^22$$

therefore the given series converges by limit comparison test with $sum e^-fracn^22$.

As an alternative by root test for $a_n=left(cosleft(frac1sqrt nright)right)^n^3$ we have

$$sqrt[n]a_n=left(cosleft(frac1sqrt nright)right)^n^2sim e^-fracn2to 0$$

share|cite|improve this answer

edited Jul 25 at 14:16

answered Jul 25 at 13:59

gimusi

65k73583

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don’t understand the downvote.
  – Szeto
  Jul 25 at 14:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve?
  – gimusi
  Jul 25 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it got downvoted, because the solution is not elementary enough?
  – Cornman
  Jul 25 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note there are two close votes on the question (for a totally silly reason, imo). It can happen that an answer gets downvoted simply bbecause people disapprove of answering "bad" questions...
  – David C. Ullrich
  Jul 25 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) math.stackexchange.com/q/2862389/505767
  – gimusi
  Jul 25 at 14:29
  

  
  
  
  

 | 
show 3 more comments

up vote
2
down vote

up vote
2
down vote

We have that

$$left(cosleft(frac1sqrt nright)right)^n^3= left(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)^n^3=e^n^3logleft(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)sim e^-fracn^22$$

therefore the given series converges by limit comparison test with $sum e^-fracn^22$.

As an alternative by root test for $a_n=left(cosleft(frac1sqrt nright)right)^n^3$ we have

$$sqrt[n]a_n=left(cosleft(frac1sqrt nright)right)^n^2sim e^-fracn2to 0$$

share|cite|improve this answer

edited Jul 25 at 14:16

answered Jul 25 at 13:59

gimusi

65k73583

We have that

$$left(cosleft(frac1sqrt nright)right)^n^3= left(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)^n^3=e^n^3logleft(1-frac12n+frac124n^2+Oleft(frac1n^3right)right)sim e^-fracn^22$$

therefore the given series converges by limit comparison test with $sum e^-fracn^22$.

As an alternative by root test for $a_n=left(cosleft(frac1sqrt nright)right)^n^3$ we have

$$sqrt[n]a_n=left(cosleft(frac1sqrt nright)right)^n^2sim e^-fracn2to 0$$

share|cite|improve this answer

edited Jul 25 at 14:16

answered Jul 25 at 13:59

gimusi

65k73583

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 14:16

edited Jul 25 at 14:16

edited Jul 25 at 14:16

answered Jul 25 at 13:59

gimusi

65k73583

answered Jul 25 at 13:59

gimusi

65k73583

answered Jul 25 at 13:59

gimusi

65k73583

65k73583

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don’t understand the downvote.
  – Szeto
  Jul 25 at 14:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve?
  – gimusi
  Jul 25 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it got downvoted, because the solution is not elementary enough?
  – Cornman
  Jul 25 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note there are two close votes on the question (for a totally silly reason, imo). It can happen that an answer gets downvoted simply bbecause people disapprove of answering "bad" questions...
  – David C. Ullrich
  Jul 25 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) math.stackexchange.com/q/2862389/505767
  – gimusi
  Jul 25 at 14:29
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I don’t understand the downvote.
  – Szeto
  Jul 25 at 14:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve?
  – gimusi
  Jul 25 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Maybe it got downvoted, because the solution is not elementary enough?
  – Cornman
  Jul 25 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note there are two close votes on the question (for a totally silly reason, imo). It can happen that an answer gets downvoted simply bbecause people disapprove of answering "bad" questions...
  – David C. Ullrich
  Jul 25 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) math.stackexchange.com/q/2862389/505767
  – gimusi
  Jul 25 at 14:29
  

  
  
  
  

1

1

I don’t understand the downvote.
– Szeto
Jul 25 at 14:09

I don’t understand the downvote.
– Szeto
Jul 25 at 14:09

@Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve?
– gimusi
Jul 25 at 14:11

@Szeto Thanks for your opinion, I was wondering the same and looking for the mistake! So you agree with that way to solve?
– gimusi
Jul 25 at 14:11

Maybe it got downvoted, because the solution is not elementary enough?
– Cornman
Jul 25 at 14:21

Maybe it got downvoted, because the solution is not elementary enough?
– Cornman
Jul 25 at 14:21

Note there are two close votes on the question (for a totally silly reason, imo). It can happen that an answer gets downvoted simply bbecause people disapprove of answering "bad" questions...
– David C. Ullrich
Jul 25 at 14:25

Note there are two close votes on the question (for a totally silly reason, imo). It can happen that an answer gets downvoted simply bbecause people disapprove of answering "bad" questions...
– David C. Ullrich
Jul 25 at 14:25

@DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) math.stackexchange.com/q/2862389/505767
– gimusi
Jul 25 at 14:29

@DavidC.Ullrich On the other han some kind of elementary questions very poorly posed are ignored, as for example (and only as an example) math.stackexchange.com/q/2862389/505767
– gimusi
Jul 25 at 14:29

 | 
show 3 more comments

up vote
2
down vote

The simplest here is to use the root test: you should find the
$$lim_ntoinftybiggl(cos^n^3!frac1sqrt nbiggr)^!tfrac1n=lim_ntoinftycos^n^2!frac1sqrt n=0.$$

Hint:

This is equivalent to showing $;lim_ntoinftyn^2logbiggl(cosdfrac1sqrt nbiggr)=-infty$, and you can use for that Taylor formula at order $2$:
$$cos u=1-fracu^22+o(u^2).$$

share|cite|improve this answer

edited Jul 25 at 14:36

answered Jul 25 at 14:24

Bernard

110k635103

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it should be $n^2$ not $1/n^2$
  – J.Dane
  Jul 25 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich: You're right. I should always check computations with a pencil and paper before posting, but I didn't. I've fixed the answer. Thank you for pointing the error!
  – Bernard
  Jul 25 at 14:39
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

The simplest here is to use the root test: you should find the
$$lim_ntoinftybiggl(cos^n^3!frac1sqrt nbiggr)^!tfrac1n=lim_ntoinftycos^n^2!frac1sqrt n=0.$$

Hint:

This is equivalent to showing $;lim_ntoinftyn^2logbiggl(cosdfrac1sqrt nbiggr)=-infty$, and you can use for that Taylor formula at order $2$:
$$cos u=1-fracu^22+o(u^2).$$

share|cite|improve this answer

edited Jul 25 at 14:36

answered Jul 25 at 14:24

Bernard

110k635103

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it should be $n^2$ not $1/n^2$
  – J.Dane
  Jul 25 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich: You're right. I should always check computations with a pencil and paper before posting, but I didn't. I've fixed the answer. Thank you for pointing the error!
  – Bernard
  Jul 25 at 14:39
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

The simplest here is to use the root test: you should find the
$$lim_ntoinftybiggl(cos^n^3!frac1sqrt nbiggr)^!tfrac1n=lim_ntoinftycos^n^2!frac1sqrt n=0.$$

Hint:

This is equivalent to showing $;lim_ntoinftyn^2logbiggl(cosdfrac1sqrt nbiggr)=-infty$, and you can use for that Taylor formula at order $2$:
$$cos u=1-fracu^22+o(u^2).$$

share|cite|improve this answer

edited Jul 25 at 14:36

answered Jul 25 at 14:24

Bernard

110k635103

The simplest here is to use the root test: you should find the
$$lim_ntoinftybiggl(cos^n^3!frac1sqrt nbiggr)^!tfrac1n=lim_ntoinftycos^n^2!frac1sqrt n=0.$$

Hint:

This is equivalent to showing $;lim_ntoinftyn^2logbiggl(cosdfrac1sqrt nbiggr)=-infty$, and you can use for that Taylor formula at order $2$:
$$cos u=1-fracu^22+o(u^2).$$

share|cite|improve this answer

edited Jul 25 at 14:36

answered Jul 25 at 14:24

Bernard

110k635103

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 14:36

edited Jul 25 at 14:36

edited Jul 25 at 14:36

answered Jul 25 at 14:24

Bernard

110k635103

answered Jul 25 at 14:24

Bernard

110k635103

answered Jul 25 at 14:24

Bernard

110k635103

110k635103

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it should be $n^2$ not $1/n^2$
  – J.Dane
  Jul 25 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich: You're right. I should always check computations with a pencil and paper before posting, but I didn't. I've fixed the answer. Thank you for pointing the error!
  – Bernard
  Jul 25 at 14:39
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I think it should be $n^2$ not $1/n^2$
  – J.Dane
  Jul 25 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidC.Ullrich: You're right. I should always check computations with a pencil and paper before posting, but I didn't. I've fixed the answer. Thank you for pointing the error!
  – Bernard
  Jul 25 at 14:39
  
  

  
  
  
  

1

1

I think it should be $n^2$ not $1/n^2$
– J.Dane
Jul 25 at 14:30

I think it should be $n^2$ not $1/n^2$
– J.Dane
Jul 25 at 14:30

@DavidC.Ullrich: You're right. I should always check computations with a pencil and paper before posting, but I didn't. I've fixed the answer. Thank you for pointing the error!
– Bernard
Jul 25 at 14:39

@DavidC.Ullrich: You're right. I should always check computations with a pencil and paper before posting, but I didn't. I've fixed the answer. Thank you for pointing the error!
– Bernard
Jul 25 at 14:39

add a comment | 

up vote
0
down vote

In style to @MarkViola's version. Here is a proof of
$$0leqcosxleq e^-fracx^22, forall xin left[0,fracpi2right]$$
using nothing but derivatives, so may be classified as elementary.
Also, $0<frac1sqrtn<fracpi2, forall n >0$. Then
$$0leqcosfrac1sqrtnleq e^-frac12nRightarrow
0leq left(cosfrac1sqrtnright)^n^3leq e^-fracn^22 tag1$$
and finally
$$0leq sumlimits_n=1 left(cosfrac1sqrtnright)^n^3leq sumlimits_n=1e^-fracn^22$$

share|cite|improve this answer

answered Jul 25 at 15:28

rtybase

8,83721333

add a comment | 

up vote
0
down vote

In style to @MarkViola's version. Here is a proof of
$$0leqcosxleq e^-fracx^22, forall xin left[0,fracpi2right]$$
using nothing but derivatives, so may be classified as elementary.
Also, $0<frac1sqrtn<fracpi2, forall n >0$. Then
$$0leqcosfrac1sqrtnleq e^-frac12nRightarrow
0leq left(cosfrac1sqrtnright)^n^3leq e^-fracn^22 tag1$$
and finally
$$0leq sumlimits_n=1 left(cosfrac1sqrtnright)^n^3leq sumlimits_n=1e^-fracn^22$$

share|cite|improve this answer

answered Jul 25 at 15:28

rtybase

8,83721333

add a comment | 

up vote
0
down vote

up vote
0
down vote

In style to @MarkViola's version. Here is a proof of
$$0leqcosxleq e^-fracx^22, forall xin left[0,fracpi2right]$$
using nothing but derivatives, so may be classified as elementary.
Also, $0<frac1sqrtn<fracpi2, forall n >0$. Then
$$0leqcosfrac1sqrtnleq e^-frac12nRightarrow
0leq left(cosfrac1sqrtnright)^n^3leq e^-fracn^22 tag1$$
and finally
$$0leq sumlimits_n=1 left(cosfrac1sqrtnright)^n^3leq sumlimits_n=1e^-fracn^22$$

share|cite|improve this answer

answered Jul 25 at 15:28

rtybase

8,83721333

In style to @MarkViola's version. Here is a proof of
$$0leqcosxleq e^-fracx^22, forall xin left[0,fracpi2right]$$
using nothing but derivatives, so may be classified as elementary.
Also, $0<frac1sqrtn<fracpi2, forall n >0$. Then
$$0leqcosfrac1sqrtnleq e^-frac12nRightarrow
0leq left(cosfrac1sqrtnright)^n^3leq e^-fracn^22 tag1$$
and finally
$$0leq sumlimits_n=1 left(cosfrac1sqrtnright)^n^3leq sumlimits_n=1e^-fracn^22$$

share|cite|improve this answer

answered Jul 25 at 15:28

rtybase

8,83721333

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 15:28

rtybase

8,83721333

answered Jul 25 at 15:28

rtybase

8,83721333

answered Jul 25 at 15:28

rtybase

8,83721333

8,83721333

add a comment | 

add a comment |