Find the interval of convergence of the following series $sum_n=1^inftyfracn^n x^nn!$

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Find the interval of convergence of the following series $$sum_n=1^inftyfracn^n x^nn!.$$

I've found the out that the series converges absolutely for $|x|<1/e$ but diverges for $|x|>1/e.$ So, when $x=1/e,$ we have

$$sum_n=1^inftyleft(fracne right)^nfrac1 n!.$$
Also, when $x=-1/e,$ we have

$$sum_n=1^infty(-1)^n left(fracne right)^nfrac1 n!.$$

My question is: how do I proceed from here?

real-analysis sequences-and-series analysis power-series

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asked Jul 14 at 20:34

Mike

74912

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you know Stirling's formula?
  – Daniel Fischer♦
  Jul 14 at 20:37
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

Find the interval of convergence of the following series $$sum_n=1^inftyfracn^n x^nn!.$$

I've found the out that the series converges absolutely for $|x|<1/e$ but diverges for $|x|>1/e.$ So, when $x=1/e,$ we have

$$sum_n=1^inftyleft(fracne right)^nfrac1 n!.$$
Also, when $x=-1/e,$ we have

$$sum_n=1^infty(-1)^n left(fracne right)^nfrac1 n!.$$

My question is: how do I proceed from here?

real-analysis sequences-and-series analysis power-series

share|cite|improve this question

asked Jul 14 at 20:34

Mike

74912

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you know Stirling's formula?
  – Daniel Fischer♦
  Jul 14 at 20:37
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

Find the interval of convergence of the following series $$sum_n=1^inftyfracn^n x^nn!.$$

I've found the out that the series converges absolutely for $|x|<1/e$ but diverges for $|x|>1/e.$ So, when $x=1/e,$ we have

$$sum_n=1^inftyleft(fracne right)^nfrac1 n!.$$
Also, when $x=-1/e,$ we have

$$sum_n=1^infty(-1)^n left(fracne right)^nfrac1 n!.$$

My question is: how do I proceed from here?

real-analysis sequences-and-series analysis power-series

share|cite|improve this question

asked Jul 14 at 20:34

Mike

74912

Find the interval of convergence of the following series $$sum_n=1^inftyfracn^n x^nn!.$$

I've found the out that the series converges absolutely for $|x|<1/e$ but diverges for $|x|>1/e.$ So, when $x=1/e,$ we have

$$sum_n=1^inftyleft(fracne right)^nfrac1 n!.$$
Also, when $x=-1/e,$ we have

$$sum_n=1^infty(-1)^n left(fracne right)^nfrac1 n!.$$

My question is: how do I proceed from here?

real-analysis sequences-and-series analysis power-series

share|cite|improve this question

asked Jul 14 at 20:34

Mike

74912

share|cite|improve this question

share|cite|improve this question

asked Jul 14 at 20:34

Mike

74912

asked Jul 14 at 20:34

Mike

74912

asked Jul 14 at 20:34

Mike

74912

74912

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you know Stirling's formula?
  – Daniel Fischer♦
  Jul 14 at 20:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Do you know Stirling's formula?
  – Daniel Fischer♦
  Jul 14 at 20:37
  

  
  
  
  

1

1

Do you know Stirling's formula?
– Daniel Fischer♦
Jul 14 at 20:37

Do you know Stirling's formula?
– Daniel Fischer♦
Jul 14 at 20:37

add a comment | 

4 Answers
4

active

oldest

votes

up vote
2
down vote



accepted

Note that $nmapsto a_nequiv(n/e)^n/n!$ is strictly decreasing. Moreover, by Stirling's approximation, $lim_nrightarrowinftya_n=0$.
Therefore, by the alternating series test, $sum(-1)^na_n$ converges.

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answered Jul 14 at 20:45

parsiad

16k32253

add a comment | 

up vote
6
down vote

HINT

By Stirling approximation

$$n! sim sqrt2 pi nleft(fracneright)^n$$

we have that

$$left(fracne right)^nfrac1 n!sim frac1sqrt2 pi n$$

share|cite|improve this answer

answered Jul 14 at 20:38

gimusi

65.4k73684

add a comment | 

up vote
1
down vote

Stirling's approximation says that$$displaystyle n!sim sqrt 2pi nleft(frac neright)^n$$therefore the first sum diverges also the 2nd sum converges since $$text2nd sumsim-1+sum_n=1^inftydfrac(-1)^nsqrt n\=-1+sum_k=1^inftydfrac1sqrt 2k-dfrac1sqrt 2k+1\=-1+sum_k=1^inftydfracsqrt 2k+1-sqrt 2ksqrt 2ksqrt 2k+1\=-1+sum_k=1^inftydfrac1sqrt 2ksqrt 2k+1dfrac1sqrt 2k+1+sqrt 2k\<-1+sum_k=1^inftydfrac12ksqrt2k+1<infty$$which is obviously convergent

share|cite|improve this answer

edited Jul 14 at 20:47

answered Jul 14 at 20:39

Mostafa Ayaz

8,6573630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sceptic about your approach with the second sum. What is the meaning of ~ in this context, this seems fishy.
  – Rumpelstiltskin
  Jul 14 at 20:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why then?......
  – Mostafa Ayaz
  Jul 14 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you compare 2 sums with alternating coefficients? I don't think you can do that... I'm pretty sure you can't
  – Rumpelstiltskin
  Jul 14 at 20:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I separated the alternating coeffs pairwise such as following$$-1+dfrac1sqrt 2-dfrac1sqrt 3+dfrac1sqrt 4-dfrac1sqrt 5+dfrac1sqrt 6-dfrac1sqrt 7+cdots=-1+(dfrac1sqrt 2-dfrac1sqrt 3)+(dfrac1sqrt 4-dfrac1sqrt 5)+(dfrac1sqrt 6-dfrac1sqrt 7)+cdots$$
  – Mostafa Ayaz
  Jul 14 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mostafa Ayaz: Please, show me the transformation you used here $~$. I couldn't follow when I got to this point.
  – Mike
  Jul 14 at 21:07
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

The radius of convergence of $sum_n=0^infty a_nx^n$ is given by $R = lim_ntoinfty left|fraca_na_n+1right|$, if this limit exists.

We have

$$left|fracfracn^nn!frac(n+1)^n+1(n+1)!right| = fracn^n(n+1)^n = frac1left(1+frac1nright)^n xrightarrowntoinfty frac1e$$

so $R = frac1e$, as you predicted.

share|cite|improve this answer

answered Jul 14 at 23:49

mechanodroid

22.3k52041

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Note that $nmapsto a_nequiv(n/e)^n/n!$ is strictly decreasing. Moreover, by Stirling's approximation, $lim_nrightarrowinftya_n=0$.
Therefore, by the alternating series test, $sum(-1)^na_n$ converges.

share|cite|improve this answer

answered Jul 14 at 20:45

parsiad

16k32253

add a comment | 

up vote
2
down vote



accepted

Note that $nmapsto a_nequiv(n/e)^n/n!$ is strictly decreasing. Moreover, by Stirling's approximation, $lim_nrightarrowinftya_n=0$.
Therefore, by the alternating series test, $sum(-1)^na_n$ converges.

share|cite|improve this answer

answered Jul 14 at 20:45

parsiad

16k32253

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Note that $nmapsto a_nequiv(n/e)^n/n!$ is strictly decreasing. Moreover, by Stirling's approximation, $lim_nrightarrowinftya_n=0$.
Therefore, by the alternating series test, $sum(-1)^na_n$ converges.

share|cite|improve this answer

answered Jul 14 at 20:45

parsiad

16k32253

Note that $nmapsto a_nequiv(n/e)^n/n!$ is strictly decreasing. Moreover, by Stirling's approximation, $lim_nrightarrowinftya_n=0$.
Therefore, by the alternating series test, $sum(-1)^na_n$ converges.

share|cite|improve this answer

answered Jul 14 at 20:45

parsiad

16k32253

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 20:45

parsiad

16k32253

answered Jul 14 at 20:45

parsiad

16k32253

answered Jul 14 at 20:45

parsiad

16k32253

16k32253

add a comment | 

add a comment | 

up vote
6
down vote

HINT

By Stirling approximation

$$n! sim sqrt2 pi nleft(fracneright)^n$$

we have that

$$left(fracne right)^nfrac1 n!sim frac1sqrt2 pi n$$

share|cite|improve this answer

answered Jul 14 at 20:38

gimusi

65.4k73684

add a comment | 

up vote
6
down vote

HINT

By Stirling approximation

$$n! sim sqrt2 pi nleft(fracneright)^n$$

we have that

$$left(fracne right)^nfrac1 n!sim frac1sqrt2 pi n$$

share|cite|improve this answer

answered Jul 14 at 20:38

gimusi

65.4k73684

add a comment | 

up vote
6
down vote

up vote
6
down vote

HINT

By Stirling approximation

$$n! sim sqrt2 pi nleft(fracneright)^n$$

we have that

$$left(fracne right)^nfrac1 n!sim frac1sqrt2 pi n$$

share|cite|improve this answer

answered Jul 14 at 20:38

gimusi

65.4k73684

HINT

By Stirling approximation

$$n! sim sqrt2 pi nleft(fracneright)^n$$

we have that

$$left(fracne right)^nfrac1 n!sim frac1sqrt2 pi n$$

share|cite|improve this answer

answered Jul 14 at 20:38

gimusi

65.4k73684

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 20:38

gimusi

65.4k73684

answered Jul 14 at 20:38

gimusi

65.4k73684

answered Jul 14 at 20:38

gimusi

65.4k73684

65.4k73684

add a comment | 

add a comment | 

up vote
1
down vote

Stirling's approximation says that$$displaystyle n!sim sqrt 2pi nleft(frac neright)^n$$therefore the first sum diverges also the 2nd sum converges since $$text2nd sumsim-1+sum_n=1^inftydfrac(-1)^nsqrt n\=-1+sum_k=1^inftydfrac1sqrt 2k-dfrac1sqrt 2k+1\=-1+sum_k=1^inftydfracsqrt 2k+1-sqrt 2ksqrt 2ksqrt 2k+1\=-1+sum_k=1^inftydfrac1sqrt 2ksqrt 2k+1dfrac1sqrt 2k+1+sqrt 2k\<-1+sum_k=1^inftydfrac12ksqrt2k+1<infty$$which is obviously convergent

share|cite|improve this answer

edited Jul 14 at 20:47

answered Jul 14 at 20:39

Mostafa Ayaz

8,6573630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sceptic about your approach with the second sum. What is the meaning of ~ in this context, this seems fishy.
  – Rumpelstiltskin
  Jul 14 at 20:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why then?......
  – Mostafa Ayaz
  Jul 14 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you compare 2 sums with alternating coefficients? I don't think you can do that... I'm pretty sure you can't
  – Rumpelstiltskin
  Jul 14 at 20:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I separated the alternating coeffs pairwise such as following$$-1+dfrac1sqrt 2-dfrac1sqrt 3+dfrac1sqrt 4-dfrac1sqrt 5+dfrac1sqrt 6-dfrac1sqrt 7+cdots=-1+(dfrac1sqrt 2-dfrac1sqrt 3)+(dfrac1sqrt 4-dfrac1sqrt 5)+(dfrac1sqrt 6-dfrac1sqrt 7)+cdots$$
  – Mostafa Ayaz
  Jul 14 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mostafa Ayaz: Please, show me the transformation you used here $~$. I couldn't follow when I got to this point.
  – Mike
  Jul 14 at 21:07
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

Stirling's approximation says that$$displaystyle n!sim sqrt 2pi nleft(frac neright)^n$$therefore the first sum diverges also the 2nd sum converges since $$text2nd sumsim-1+sum_n=1^inftydfrac(-1)^nsqrt n\=-1+sum_k=1^inftydfrac1sqrt 2k-dfrac1sqrt 2k+1\=-1+sum_k=1^inftydfracsqrt 2k+1-sqrt 2ksqrt 2ksqrt 2k+1\=-1+sum_k=1^inftydfrac1sqrt 2ksqrt 2k+1dfrac1sqrt 2k+1+sqrt 2k\<-1+sum_k=1^inftydfrac12ksqrt2k+1<infty$$which is obviously convergent

share|cite|improve this answer

edited Jul 14 at 20:47

answered Jul 14 at 20:39

Mostafa Ayaz

8,6573630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sceptic about your approach with the second sum. What is the meaning of ~ in this context, this seems fishy.
  – Rumpelstiltskin
  Jul 14 at 20:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why then?......
  – Mostafa Ayaz
  Jul 14 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you compare 2 sums with alternating coefficients? I don't think you can do that... I'm pretty sure you can't
  – Rumpelstiltskin
  Jul 14 at 20:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I separated the alternating coeffs pairwise such as following$$-1+dfrac1sqrt 2-dfrac1sqrt 3+dfrac1sqrt 4-dfrac1sqrt 5+dfrac1sqrt 6-dfrac1sqrt 7+cdots=-1+(dfrac1sqrt 2-dfrac1sqrt 3)+(dfrac1sqrt 4-dfrac1sqrt 5)+(dfrac1sqrt 6-dfrac1sqrt 7)+cdots$$
  – Mostafa Ayaz
  Jul 14 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mostafa Ayaz: Please, show me the transformation you used here $~$. I couldn't follow when I got to this point.
  – Mike
  Jul 14 at 21:07
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

up vote
1
down vote

Stirling's approximation says that$$displaystyle n!sim sqrt 2pi nleft(frac neright)^n$$therefore the first sum diverges also the 2nd sum converges since $$text2nd sumsim-1+sum_n=1^inftydfrac(-1)^nsqrt n\=-1+sum_k=1^inftydfrac1sqrt 2k-dfrac1sqrt 2k+1\=-1+sum_k=1^inftydfracsqrt 2k+1-sqrt 2ksqrt 2ksqrt 2k+1\=-1+sum_k=1^inftydfrac1sqrt 2ksqrt 2k+1dfrac1sqrt 2k+1+sqrt 2k\<-1+sum_k=1^inftydfrac12ksqrt2k+1<infty$$which is obviously convergent

share|cite|improve this answer

edited Jul 14 at 20:47

answered Jul 14 at 20:39

Mostafa Ayaz

8,6573630

Stirling's approximation says that$$displaystyle n!sim sqrt 2pi nleft(frac neright)^n$$therefore the first sum diverges also the 2nd sum converges since $$text2nd sumsim-1+sum_n=1^inftydfrac(-1)^nsqrt n\=-1+sum_k=1^inftydfrac1sqrt 2k-dfrac1sqrt 2k+1\=-1+sum_k=1^inftydfracsqrt 2k+1-sqrt 2ksqrt 2ksqrt 2k+1\=-1+sum_k=1^inftydfrac1sqrt 2ksqrt 2k+1dfrac1sqrt 2k+1+sqrt 2k\<-1+sum_k=1^inftydfrac12ksqrt2k+1<infty$$which is obviously convergent

share|cite|improve this answer

edited Jul 14 at 20:47

answered Jul 14 at 20:39

Mostafa Ayaz

8,6573630

share|cite|improve this answer

share|cite|improve this answer

edited Jul 14 at 20:47

edited Jul 14 at 20:47

edited Jul 14 at 20:47

answered Jul 14 at 20:39

Mostafa Ayaz

8,6573630

answered Jul 14 at 20:39

Mostafa Ayaz

8,6573630

answered Jul 14 at 20:39

Mostafa Ayaz

8,6573630

8,6573630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sceptic about your approach with the second sum. What is the meaning of ~ in this context, this seems fishy.
  – Rumpelstiltskin
  Jul 14 at 20:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why then?......
  – Mostafa Ayaz
  Jul 14 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you compare 2 sums with alternating coefficients? I don't think you can do that... I'm pretty sure you can't
  – Rumpelstiltskin
  Jul 14 at 20:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I separated the alternating coeffs pairwise such as following$$-1+dfrac1sqrt 2-dfrac1sqrt 3+dfrac1sqrt 4-dfrac1sqrt 5+dfrac1sqrt 6-dfrac1sqrt 7+cdots=-1+(dfrac1sqrt 2-dfrac1sqrt 3)+(dfrac1sqrt 4-dfrac1sqrt 5)+(dfrac1sqrt 6-dfrac1sqrt 7)+cdots$$
  – Mostafa Ayaz
  Jul 14 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mostafa Ayaz: Please, show me the transformation you used here $~$. I couldn't follow when I got to this point.
  – Mike
  Jul 14 at 21:07
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm sceptic about your approach with the second sum. What is the meaning of ~ in this context, this seems fishy.
  – Rumpelstiltskin
  Jul 14 at 20:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why then?......
  – Mostafa Ayaz
  Jul 14 at 20:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you compare 2 sums with alternating coefficients? I don't think you can do that... I'm pretty sure you can't
  – Rumpelstiltskin
  Jul 14 at 20:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I separated the alternating coeffs pairwise such as following$$-1+dfrac1sqrt 2-dfrac1sqrt 3+dfrac1sqrt 4-dfrac1sqrt 5+dfrac1sqrt 6-dfrac1sqrt 7+cdots=-1+(dfrac1sqrt 2-dfrac1sqrt 3)+(dfrac1sqrt 4-dfrac1sqrt 5)+(dfrac1sqrt 6-dfrac1sqrt 7)+cdots$$
  – Mostafa Ayaz
  Jul 14 at 21:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mostafa Ayaz: Please, show me the transformation you used here $~$. I couldn't follow when I got to this point.
  – Mike
  Jul 14 at 21:07
  

  
  
  
  

I'm sceptic about your approach with the second sum. What is the meaning of ~ in this context, this seems fishy.
– Rumpelstiltskin
Jul 14 at 20:50

I'm sceptic about your approach with the second sum. What is the meaning of ~ in this context, this seems fishy.
– Rumpelstiltskin
Jul 14 at 20:50

Why then?......
– Mostafa Ayaz
Jul 14 at 20:51

Why then?......
– Mostafa Ayaz
Jul 14 at 20:51

Did you compare 2 sums with alternating coefficients? I don't think you can do that... I'm pretty sure you can't
– Rumpelstiltskin
Jul 14 at 20:57

Did you compare 2 sums with alternating coefficients? I don't think you can do that... I'm pretty sure you can't
– Rumpelstiltskin
Jul 14 at 20:57

I separated the alternating coeffs pairwise such as following$$-1+dfrac1sqrt 2-dfrac1sqrt 3+dfrac1sqrt 4-dfrac1sqrt 5+dfrac1sqrt 6-dfrac1sqrt 7+cdots=-1+(dfrac1sqrt 2-dfrac1sqrt 3)+(dfrac1sqrt 4-dfrac1sqrt 5)+(dfrac1sqrt 6-dfrac1sqrt 7)+cdots$$
– Mostafa Ayaz
Jul 14 at 21:00

I separated the alternating coeffs pairwise such as following$$-1+dfrac1sqrt 2-dfrac1sqrt 3+dfrac1sqrt 4-dfrac1sqrt 5+dfrac1sqrt 6-dfrac1sqrt 7+cdots=-1+(dfrac1sqrt 2-dfrac1sqrt 3)+(dfrac1sqrt 4-dfrac1sqrt 5)+(dfrac1sqrt 6-dfrac1sqrt 7)+cdots$$
– Mostafa Ayaz
Jul 14 at 21:00

@Mostafa Ayaz: Please, show me the transformation you used here $~$. I couldn't follow when I got to this point.
– Mike
Jul 14 at 21:07

@Mostafa Ayaz: Please, show me the transformation you used here $~$. I couldn't follow when I got to this point.
– Mike
Jul 14 at 21:07

 | 
show 2 more comments

up vote
1
down vote

The radius of convergence of $sum_n=0^infty a_nx^n$ is given by $R = lim_ntoinfty left|fraca_na_n+1right|$, if this limit exists.

We have

$$left|fracfracn^nn!frac(n+1)^n+1(n+1)!right| = fracn^n(n+1)^n = frac1left(1+frac1nright)^n xrightarrowntoinfty frac1e$$

so $R = frac1e$, as you predicted.

share|cite|improve this answer

answered Jul 14 at 23:49

mechanodroid

22.3k52041

add a comment | 

up vote
1
down vote

The radius of convergence of $sum_n=0^infty a_nx^n$ is given by $R = lim_ntoinfty left|fraca_na_n+1right|$, if this limit exists.

We have

$$left|fracfracn^nn!frac(n+1)^n+1(n+1)!right| = fracn^n(n+1)^n = frac1left(1+frac1nright)^n xrightarrowntoinfty frac1e$$

so $R = frac1e$, as you predicted.

share|cite|improve this answer

answered Jul 14 at 23:49

mechanodroid

22.3k52041

add a comment | 

up vote
1
down vote

up vote
1
down vote

The radius of convergence of $sum_n=0^infty a_nx^n$ is given by $R = lim_ntoinfty left|fraca_na_n+1right|$, if this limit exists.

We have

$$left|fracfracn^nn!frac(n+1)^n+1(n+1)!right| = fracn^n(n+1)^n = frac1left(1+frac1nright)^n xrightarrowntoinfty frac1e$$

so $R = frac1e$, as you predicted.

share|cite|improve this answer

answered Jul 14 at 23:49

mechanodroid

22.3k52041

The radius of convergence of $sum_n=0^infty a_nx^n$ is given by $R = lim_ntoinfty left|fraca_na_n+1right|$, if this limit exists.

We have

$$left|fracfracn^nn!frac(n+1)^n+1(n+1)!right| = fracn^n(n+1)^n = frac1left(1+frac1nright)^n xrightarrowntoinfty frac1e$$

so $R = frac1e$, as you predicted.

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answered Jul 14 at 23:49

mechanodroid

22.3k52041

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share|cite|improve this answer

answered Jul 14 at 23:49

mechanodroid

22.3k52041

answered Jul 14 at 23:49

mechanodroid

22.3k52041

answered Jul 14 at 23:49

mechanodroid

22.3k52041

22.3k52041

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