Following series convergent/divergent

**utopiaNow** · June 10th 2009, 07:24 PM

Here are the 2 series where their nth terms are shown:

$ c_n = (-1)^n \frac{(n+1)^n}{n^n} $

$ d_n = \frac{(n+1)^n}{n^{n + 1}} $

Attempted solution:
So I know about all the tests: monotone convergence criterion, comparison test, root test, ratio test, and alternating series test.

For $c_n$ I tried to use the alternating series test however it's neither nonincreasing nor clear if $\displaystyle\lim_{n\to\infty}c_n = 0$ .

For $d_n$ my hunch is to say that since $n \geq 1 \Rightarrow 1^n \leq \frac{(n+1)^n}{n^{n + 1}}$ . Therefore by the comparison test, since $1^n$ diverges so does $d_n$ .

**TheEmptySet** · June 10th 2009, 08:06 PM

Originally Posted by utopiaNow

Here are the 2 series where their nth terms are shown:

$ c_n = (-1)^n \frac{(n+1)^n}{n^n} $

$ d_n = \frac{(n+1)^n}{n^{n + 1}} $

Attempted solution:
So I know about all the tests: monotone convergence criterion, comparison test, root test, ratio test, and alternating series test.

For $c_n$ I tried to use the alternating series test however it's neither nonincreasing nor clear if $\displaystyle\lim_{n\to\infty}c_n = 0$ .

For $d_n$ my hunch is to say that since $n \geq 1 \Rightarrow 1^n \leq \frac{(n+1)^n}{n^{n + 1}}$ . Therefore by the comparison test, since $1^n$ diverges so does $d_n$ .

For the first notice that

$\frac{(n+1)^n}{n^n}=\left( \frac{n+1}{n} \right)^n =\left( 1+\frac{1}{n} \right)^n$

This is a very famous limit

$\lim_{n \to \infty}\left( 1+\frac{1}{n} \right)^n =e$

So it is divergent

I hope this helps

**putnam120** · June 11th 2009, 03:38 PM

$d_n=\frac{1}{n}\left(1+\frac{1}{n}\right)^n\longri ghtarrow{0}$ as $n\to\infty$ . To see that $\sum d_n$ diverges just notice that $\frac{1}{n}<\frac{(n+1)^n}{n^{n+1}}$ .

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