1. ## Following series convergent/divergent

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

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

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

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

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

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

2. Originally Posted by utopiaNow
Here are the 2 series where their nth terms are shown:

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

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

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

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

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

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

This is a very famous limit

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

So it is divergent

I hope this helps

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