Is the series $sum_k=1^infty frac1k^3/2 $ convergent or divergent? [closed]

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$sum_k=1^infty frac1k^3/2 $ should be between the harmonic series $sum_k=1^infty frac1k $ which diverges and $sum_k=1^infty frac1k^2 $ which converges.

So I am not sure how should I approach this problem.

real-analysis power-series divergent-series

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edited Jul 22 at 0:10

amWhy

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asked Jul 21 at 23:21

emily

493

closed as off-topic by amWhy, GEdgar, T. Bongers, Adrian Keister, Leucippus Jul 22 at 2:15

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." – amWhy, GEdgar, T. Bongers, Adrian Keister, Leucippus

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

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So you know that the harmonic series diverges and the reciprocals of the squares converges. How were those two facts established? Then use the same technique on your series.
  – B. Goddard
  Jul 21 at 23:35
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

$sum_k=1^infty frac1k^3/2 $ should be between the harmonic series $sum_k=1^infty frac1k $ which diverges and $sum_k=1^infty frac1k^2 $ which converges.

So I am not sure how should I approach this problem.

real-analysis power-series divergent-series

share|cite|improve this question

edited Jul 22 at 0:10

amWhy

189k25219431

asked Jul 21 at 23:21

emily

493

closed as off-topic by amWhy, GEdgar, T. Bongers, Adrian Keister, Leucippus Jul 22 at 2:15

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." – amWhy, GEdgar, T. Bongers, Adrian Keister, Leucippus

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

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So you know that the harmonic series diverges and the reciprocals of the squares converges. How were those two facts established? Then use the same technique on your series.
  – B. Goddard
  Jul 21 at 23:35
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

$sum_k=1^infty frac1k^3/2 $ should be between the harmonic series $sum_k=1^infty frac1k $ which diverges and $sum_k=1^infty frac1k^2 $ which converges.

So I am not sure how should I approach this problem.

real-analysis power-series divergent-series

share|cite|improve this question

edited Jul 22 at 0:10

amWhy

189k25219431

asked Jul 21 at 23:21

emily

493

$sum_k=1^infty frac1k^3/2 $ should be between the harmonic series $sum_k=1^infty frac1k $ which diverges and $sum_k=1^infty frac1k^2 $ which converges.

So I am not sure how should I approach this problem.

real-analysis power-series divergent-series

share|cite|improve this question

edited Jul 22 at 0:10

amWhy

189k25219431

asked Jul 21 at 23:21

emily

493

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 0:10

amWhy

189k25219431

edited Jul 22 at 0:10

amWhy

189k25219431

edited Jul 22 at 0:10

amWhy

189k25219431

189k25219431

asked Jul 21 at 23:21

emily

493

asked Jul 21 at 23:21

emily

493

asked Jul 21 at 23:21

emily

493

493

closed as off-topic by amWhy, GEdgar, T. Bongers, Adrian Keister, Leucippus Jul 22 at 2:15

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." – amWhy, GEdgar, T. Bongers, Adrian Keister, Leucippus

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

closed as off-topic by amWhy, GEdgar, T. Bongers, Adrian Keister, Leucippus Jul 22 at 2:15

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." – amWhy, GEdgar, T. Bongers, Adrian Keister, Leucippus

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

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So you know that the harmonic series diverges and the reciprocals of the squares converges. How were those two facts established? Then use the same technique on your series.
  – B. Goddard
  Jul 21 at 23:35
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So you know that the harmonic series diverges and the reciprocals of the squares converges. How were those two facts established? Then use the same technique on your series.
  – B. Goddard
  Jul 21 at 23:35
  

  
  
  
  

4

4

So you know that the harmonic series diverges and the reciprocals of the squares converges. How were those two facts established? Then use the same technique on your series.
– B. Goddard
Jul 21 at 23:35

So you know that the harmonic series diverges and the reciprocals of the squares converges. How were those two facts established? Then use the same technique on your series.
– B. Goddard
Jul 21 at 23:35

add a comment | 

4 Answers
4

active

oldest

votes

up vote
3
down vote

Given $$sum_k=1^infty frac1k^3/2$$

Use $p$-series test:

Given a series of the form $displaystylesum_n=1^inftyfrac1n^p,;$ if $p>1,$ then series converges.

Here $p=dfrac32>1$.

Therefore, the series converges.

share|cite|improve this answer

edited Jul 21 at 23:34

amWhy

189k25219431

answered Jul 21 at 23:25

Key Flex

4,318424

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm guessing the OP does not know the $p$-series test.
  – GEdgar
  Jul 21 at 23:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Unfortunately, I don't think this is a good answer. If the asker knows the $p$-series test then I doubt they would ask this question; without knowing anything about their background I'm not sure this is helpful.
  – T. Bongers
  Jul 22 at 0:30
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This doesn't even come close to an appropriate reply to the question.
  – Mark Viola
  Jul 22 at 1:17
  

  
  
  
  

add a comment | 

up vote
2
down vote

Note that using creative telescoping we have

$$beginalign
2left(1-frac1(K+1)^1/2right)&=2sum_k=1^K left(frac1k^1/2-frac1(k+1)^1/2right)\\
&=2sum_k=1^K left(frac1k^1/2(k+1)^1/2(k^1/2+(k+1)^1/2)right)\\
&ge sum_k=1^K frac1(k+1)^3/2\\
&=sum_k=2^K+1frac1k^3/2tag1
endalign$$

Hence, we see from $(1)$ that

$$sum_k=1^Kfrac1k^3/2le 1-frac1(K+1)^3/2+2left(1-frac1(K+1)^1/2right)tag 2$$

The right-hand side approaches $3$ as $Kto infty$. Since the partial sums on the left-hand side of $(2)$ are increasing and bounded, the series converges.

share|cite|improve this answer

answered Jul 22 at 0:46

Mark Viola

126k1172167

add a comment | 

up vote
1
down vote

The idea is to compare it with an integral namely $$ int _1 ^infty frac 1x^3/2 dx $$

Since the integral converges the series converges as well.

This is the idea behind the "P-Test" which indicates $$sum _1^infty frac 1n^p$$ converges for $p>1$ and diverges otherwise.

share|cite|improve this answer

answered Jul 21 at 23:36

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
0
down vote

Use the Cauchy condensation test.

The series

$$sum_n=1^infty 2^n frac1(2^n)^3/2 = sum_n=1^infty frac1(sqrt2)^n$$

converges because it is a geometric series. Hence $sum_n=1^infty frac1n^3/2$ also converges.

share|cite|improve this answer

answered Jul 21 at 23:40

mechanodroid

22.2k52041

add a comment | 

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Given $$sum_k=1^infty frac1k^3/2$$

Use $p$-series test:

Given a series of the form $displaystylesum_n=1^inftyfrac1n^p,;$ if $p>1,$ then series converges.

Here $p=dfrac32>1$.

Therefore, the series converges.

share|cite|improve this answer

edited Jul 21 at 23:34

amWhy

189k25219431

answered Jul 21 at 23:25

Key Flex

4,318424

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm guessing the OP does not know the $p$-series test.
  – GEdgar
  Jul 21 at 23:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Unfortunately, I don't think this is a good answer. If the asker knows the $p$-series test then I doubt they would ask this question; without knowing anything about their background I'm not sure this is helpful.
  – T. Bongers
  Jul 22 at 0:30
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This doesn't even come close to an appropriate reply to the question.
  – Mark Viola
  Jul 22 at 1:17
  

  
  
  
  

add a comment | 

up vote
3
down vote

Given $$sum_k=1^infty frac1k^3/2$$

Use $p$-series test:

Given a series of the form $displaystylesum_n=1^inftyfrac1n^p,;$ if $p>1,$ then series converges.

Here $p=dfrac32>1$.

Therefore, the series converges.

share|cite|improve this answer

edited Jul 21 at 23:34

amWhy

189k25219431

answered Jul 21 at 23:25

Key Flex

4,318424

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm guessing the OP does not know the $p$-series test.
  – GEdgar
  Jul 21 at 23:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Unfortunately, I don't think this is a good answer. If the asker knows the $p$-series test then I doubt they would ask this question; without knowing anything about their background I'm not sure this is helpful.
  – T. Bongers
  Jul 22 at 0:30
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This doesn't even come close to an appropriate reply to the question.
  – Mark Viola
  Jul 22 at 1:17
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Given $$sum_k=1^infty frac1k^3/2$$

Use $p$-series test:

Given a series of the form $displaystylesum_n=1^inftyfrac1n^p,;$ if $p>1,$ then series converges.

Here $p=dfrac32>1$.

Therefore, the series converges.

share|cite|improve this answer

edited Jul 21 at 23:34

amWhy

189k25219431

answered Jul 21 at 23:25

Key Flex

4,318424

Given $$sum_k=1^infty frac1k^3/2$$

Use $p$-series test:

Given a series of the form $displaystylesum_n=1^inftyfrac1n^p,;$ if $p>1,$ then series converges.

Here $p=dfrac32>1$.

Therefore, the series converges.

share|cite|improve this answer

edited Jul 21 at 23:34

amWhy

189k25219431

answered Jul 21 at 23:25

Key Flex

4,318424

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 23:34

amWhy

189k25219431

edited Jul 21 at 23:34

amWhy

189k25219431

edited Jul 21 at 23:34

amWhy

189k25219431

189k25219431

answered Jul 21 at 23:25

Key Flex

4,318424

answered Jul 21 at 23:25

Key Flex

4,318424

answered Jul 21 at 23:25

Key Flex

4,318424

4,318424

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm guessing the OP does not know the $p$-series test.
  – GEdgar
  Jul 21 at 23:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Unfortunately, I don't think this is a good answer. If the asker knows the $p$-series test then I doubt they would ask this question; without knowing anything about their background I'm not sure this is helpful.
  – T. Bongers
  Jul 22 at 0:30
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This doesn't even come close to an appropriate reply to the question.
  – Mark Viola
  Jul 22 at 1:17
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm guessing the OP does not know the $p$-series test.
  – GEdgar
  Jul 21 at 23:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Unfortunately, I don't think this is a good answer. If the asker knows the $p$-series test then I doubt they would ask this question; without knowing anything about their background I'm not sure this is helpful.
  – T. Bongers
  Jul 22 at 0:30
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  This doesn't even come close to an appropriate reply to the question.
  – Mark Viola
  Jul 22 at 1:17
  

  
  
  
  

1

1

I'm guessing the OP does not know the $p$-series test.
– GEdgar
Jul 21 at 23:57

I'm guessing the OP does not know the $p$-series test.
– GEdgar
Jul 21 at 23:57

1

1

Unfortunately, I don't think this is a good answer. If the asker knows the $p$-series test then I doubt they would ask this question; without knowing anything about their background I'm not sure this is helpful.
– T. Bongers
Jul 22 at 0:30

Unfortunately, I don't think this is a good answer. If the asker knows the $p$-series test then I doubt they would ask this question; without knowing anything about their background I'm not sure this is helpful.
– T. Bongers
Jul 22 at 0:30

2

2

This doesn't even come close to an appropriate reply to the question.
– Mark Viola
Jul 22 at 1:17

This doesn't even come close to an appropriate reply to the question.
– Mark Viola
Jul 22 at 1:17

add a comment | 

up vote
2
down vote

Note that using creative telescoping we have

$$beginalign
2left(1-frac1(K+1)^1/2right)&=2sum_k=1^K left(frac1k^1/2-frac1(k+1)^1/2right)\\
&=2sum_k=1^K left(frac1k^1/2(k+1)^1/2(k^1/2+(k+1)^1/2)right)\\
&ge sum_k=1^K frac1(k+1)^3/2\\
&=sum_k=2^K+1frac1k^3/2tag1
endalign$$

Hence, we see from $(1)$ that

$$sum_k=1^Kfrac1k^3/2le 1-frac1(K+1)^3/2+2left(1-frac1(K+1)^1/2right)tag 2$$

The right-hand side approaches $3$ as $Kto infty$. Since the partial sums on the left-hand side of $(2)$ are increasing and bounded, the series converges.

share|cite|improve this answer

answered Jul 22 at 0:46

Mark Viola

126k1172167

add a comment | 

up vote
2
down vote

Note that using creative telescoping we have

$$beginalign
2left(1-frac1(K+1)^1/2right)&=2sum_k=1^K left(frac1k^1/2-frac1(k+1)^1/2right)\\
&=2sum_k=1^K left(frac1k^1/2(k+1)^1/2(k^1/2+(k+1)^1/2)right)\\
&ge sum_k=1^K frac1(k+1)^3/2\\
&=sum_k=2^K+1frac1k^3/2tag1
endalign$$

Hence, we see from $(1)$ that

$$sum_k=1^Kfrac1k^3/2le 1-frac1(K+1)^3/2+2left(1-frac1(K+1)^1/2right)tag 2$$

The right-hand side approaches $3$ as $Kto infty$. Since the partial sums on the left-hand side of $(2)$ are increasing and bounded, the series converges.

share|cite|improve this answer

answered Jul 22 at 0:46

Mark Viola

126k1172167

add a comment | 

up vote
2
down vote

up vote
2
down vote

Note that using creative telescoping we have

$$beginalign
2left(1-frac1(K+1)^1/2right)&=2sum_k=1^K left(frac1k^1/2-frac1(k+1)^1/2right)\\
&=2sum_k=1^K left(frac1k^1/2(k+1)^1/2(k^1/2+(k+1)^1/2)right)\\
&ge sum_k=1^K frac1(k+1)^3/2\\
&=sum_k=2^K+1frac1k^3/2tag1
endalign$$

Hence, we see from $(1)$ that

$$sum_k=1^Kfrac1k^3/2le 1-frac1(K+1)^3/2+2left(1-frac1(K+1)^1/2right)tag 2$$

The right-hand side approaches $3$ as $Kto infty$. Since the partial sums on the left-hand side of $(2)$ are increasing and bounded, the series converges.

share|cite|improve this answer

answered Jul 22 at 0:46

Mark Viola

126k1172167

Note that using creative telescoping we have

$$beginalign
2left(1-frac1(K+1)^1/2right)&=2sum_k=1^K left(frac1k^1/2-frac1(k+1)^1/2right)\\
&=2sum_k=1^K left(frac1k^1/2(k+1)^1/2(k^1/2+(k+1)^1/2)right)\\
&ge sum_k=1^K frac1(k+1)^3/2\\
&=sum_k=2^K+1frac1k^3/2tag1
endalign$$

Hence, we see from $(1)$ that

$$sum_k=1^Kfrac1k^3/2le 1-frac1(K+1)^3/2+2left(1-frac1(K+1)^1/2right)tag 2$$

The right-hand side approaches $3$ as $Kto infty$. Since the partial sums on the left-hand side of $(2)$ are increasing and bounded, the series converges.

share|cite|improve this answer

answered Jul 22 at 0:46

Mark Viola

126k1172167

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 0:46

Mark Viola

126k1172167

answered Jul 22 at 0:46

Mark Viola

126k1172167

answered Jul 22 at 0:46

Mark Viola

126k1172167

126k1172167

add a comment | 

add a comment | 

up vote
1
down vote

The idea is to compare it with an integral namely $$ int _1 ^infty frac 1x^3/2 dx $$

Since the integral converges the series converges as well.

This is the idea behind the "P-Test" which indicates $$sum _1^infty frac 1n^p$$ converges for $p>1$ and diverges otherwise.

share|cite|improve this answer

answered Jul 21 at 23:36

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
1
down vote

The idea is to compare it with an integral namely $$ int _1 ^infty frac 1x^3/2 dx $$

Since the integral converges the series converges as well.

This is the idea behind the "P-Test" which indicates $$sum _1^infty frac 1n^p$$ converges for $p>1$ and diverges otherwise.

share|cite|improve this answer

answered Jul 21 at 23:36

Mohammad Riazi-Kermani

27.5k41852

add a comment | 

up vote
1
down vote

up vote
1
down vote

The idea is to compare it with an integral namely $$ int _1 ^infty frac 1x^3/2 dx $$

Since the integral converges the series converges as well.

This is the idea behind the "P-Test" which indicates $$sum _1^infty frac 1n^p$$ converges for $p>1$ and diverges otherwise.

share|cite|improve this answer

answered Jul 21 at 23:36

Mohammad Riazi-Kermani

27.5k41852

The idea is to compare it with an integral namely $$ int _1 ^infty frac 1x^3/2 dx $$

Since the integral converges the series converges as well.

This is the idea behind the "P-Test" which indicates $$sum _1^infty frac 1n^p$$ converges for $p>1$ and diverges otherwise.

share|cite|improve this answer

answered Jul 21 at 23:36

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 23:36

Mohammad Riazi-Kermani

27.5k41852

answered Jul 21 at 23:36

Mohammad Riazi-Kermani

27.5k41852

answered Jul 21 at 23:36

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

add a comment | 

add a comment | 

up vote
0
down vote

Use the Cauchy condensation test.

The series

$$sum_n=1^infty 2^n frac1(2^n)^3/2 = sum_n=1^infty frac1(sqrt2)^n$$

converges because it is a geometric series. Hence $sum_n=1^infty frac1n^3/2$ also converges.

share|cite|improve this answer

answered Jul 21 at 23:40

mechanodroid

22.2k52041

add a comment | 

up vote
0
down vote

Use the Cauchy condensation test.

The series

$$sum_n=1^infty 2^n frac1(2^n)^3/2 = sum_n=1^infty frac1(sqrt2)^n$$

converges because it is a geometric series. Hence $sum_n=1^infty frac1n^3/2$ also converges.

share|cite|improve this answer

answered Jul 21 at 23:40

mechanodroid

22.2k52041

add a comment | 

up vote
0
down vote

up vote
0
down vote

Use the Cauchy condensation test.

The series

$$sum_n=1^infty 2^n frac1(2^n)^3/2 = sum_n=1^infty frac1(sqrt2)^n$$

converges because it is a geometric series. Hence $sum_n=1^infty frac1n^3/2$ also converges.

share|cite|improve this answer

answered Jul 21 at 23:40

mechanodroid

22.2k52041

Use the Cauchy condensation test.

The series

$$sum_n=1^infty 2^n frac1(2^n)^3/2 = sum_n=1^infty frac1(sqrt2)^n$$

converges because it is a geometric series. Hence $sum_n=1^infty frac1n^3/2$ also converges.

share|cite|improve this answer

answered Jul 21 at 23:40

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 23:40

mechanodroid

22.2k52041

answered Jul 21 at 23:40

mechanodroid

22.2k52041

answered Jul 21 at 23:40

mechanodroid

22.2k52041

22.2k52041

add a comment | 

add a comment |