# Help in infinite series problem

1. May 28, 2014

### danunicamp

1. The problem statement, all variables and given/known data

Prove that:
[n+1 / n^2 + (n+1)^2 / n^3 + ... + (n+1)^n / n ^ (n+1) -> e-1

2. Relevant equations

I have been trying it for couple of days. Tried to work the terms, natural log it all, use the byniomial theory but i can´t get to the right answer.

3. The attempt at a solution

Working the terms:

n+1/n^2 = n( 1+1/n) / n^2 = (1+1/n)/n
(n+1)^2/n^3 = n^2 (1+1/n) ^2 / n^3 = (1+1/n)^2/n

Ok, so at n:

(n+1)^n / n^(n+1) = n^n (1+1/n) ^ n / n ^(n+1) = (1+1/n) ^n / n

[(1+1/n) + (1+1/n)^2 +... (1+1/n)^n] / n

Adding the sequence: ( a + ar^2 + ar^3 + ar^n = a / 1 - r)

(1/n) / [1 - (1+1/n)] = (1/n) / -1/n = -1

Second try using logarithm got to:

1 - [ ln (1+1/n)^n / ln n ] and couldn,t progress further

Thank you very much in advance for any help provided.

2. May 28, 2014

### jbunniii

Looks OK up to here.

No, you're only adding finitely many terms. If we put $a = 1 + 1/n$, you have
$$\frac{1}{n}(a + a^2 + a^3 + \cdots + a^n) = \frac{a}{n}(1 + a + a^2 + \cdots + a^{n-1}) = \frac{a}{n}\left(\frac{1 - a^n}{1-a}\right)$$
Now plug in $a = 1 + 1/n$ and simplify.

3. May 28, 2014

### danunicamp

Thank you very very much.

I understand it now and finally solved it.

Thank you again!!