Help with sum of infinite series using the root test.

  • #1
Homework Statement

Does it converge or diverge?

Sum n=0 to infinity : (n/(n+1))^(n^2)

The attempt at a solution

I know I need to use the root test.
But what I get is ...

Limit to infinity : (n/(n+1))^n

It seems that the n/(n+1) would go to 1 because when you multiply the top and bottom by n^-1 to get 1/(1+1/n) the 1/n goes to zero and you have 1/1. 1^n is 1, and according to the root test, it is inconclusive. However, the series converges and apparently the limit goes to 1/e, but I'm not sure how.

Also, does the latex syntax not work in preview mode or something?
 

Answers and Replies

  • #2
vela
Staff Emeritus
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To evaluate

[tex]\lim_{n\rightarrow\infty} (1+1/n)^n[/tex]

try finding the limit of the log of the function.
 
  • #3
To evaluate

[tex]\lim_{n\rightarrow\infty} (1+1/n)^n[/tex]

try finding the limit of the log of the function.

I have no idea how.
 
  • #4
HallsofIvy
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Are you saying that you do not know how to find a logarithm??

If [itex]y= (1+ 1/n)^n[/itex], then [itex]log(y)= n log(1+ 1/n)[/itex]. What is the limit of that as n goes to infinity? Perhaps L'Hopital's rule or writing log(1+ 1/n) as a Taylor's series will help.
 
  • #5
24
0
Homework Statement

Does it converge or diverge?

Sum n=0 to infinity : (n/(n+1))^(n^2)

The attempt at a solution

I know I need to use the root test.
But what I get is ...

Limit to infinity : (n/(n+1))^n

It seems that the n/(n+1) would go to 1 because when you multiply the top and bottom by n^-1 to get 1/(1+1/n) the 1/n goes to zero and you have 1/1. 1^n is 1, and according to the root test, it is inconclusive. However, the series converges and apparently the limit goes to 1/e, but I'm not sure how.

Also, does the latex syntax not work in preview mode or something?

[tex]\lim_{n\to\infty} \left( \frac{n}{n+1} \right)^n = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^{-n} = ?[/tex]

Recall:
[tex]\lim_{n\to\infty} \left( 1 + \frac{a}{n^k} \right)^{bn^k} = e^{ab}[/tex]
 

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