1. ## Infinite Series Problems

Hey guys, I have some infinite series problems I need my solutions checked to.

Problem 1. Use the ratio test to determine whether the series converges or diverges. Explain how the test permits you to draw your conclusion.

$\Sigma^{\infty}_{n = 1}\frac{n^n}{n!}$

My Solution

$\lim_{n \to \infty}\frac{n^{n + 1}}{(n+1)n!}*\frac{n!}{n^n} = \lim_{n \to \infty}\frac{n}{n + 1} = 1$

Since $\lim_{n \to \infty}\frac{\frac{n^{n + 1}}{(n + 1)!}}{\frac{n^n}{n!}} = 1$, the ratio test fails.

Here, I have trouble thinking my solution is right because the directions make no mention of the test failing. Is my solution correct?

For problems 2 and 3 determine whether the series converges absolutely, converges conditionally, or diverges. Explain how you drew your conclusion in each case.

Problem 2.

$\Sigma^{\infty}_{n = 1}(-1)^{n}\frac{e^n}{n!}$

Problem 3.

$\Sigma^{\infty}_{n = 1}(-1)^{n}\frac{1}{\sqrt{2n + 1}}$

I don't know how to do either of these problems, help please!

Thanks in advance for all help!

2. Problem one remember that (n+1)! is (n+1)n! so that cancels out the other n!.
(N+1)^n+1 x n! = (n+1)^n+1/(n+1)..= [(n+1)/n]^n=e
(n+1)! x N^n

e is larger than 1 there for diverges

3. For problem 2 and 3 use ratio test (n+1)/n and you will find out if it is absouletly convergent or not

4. Originally Posted by mturner07
Hey guys, I have some infinite series problems I need my solutions checked to.

Problem 1. Use the ratio test to determine whether the series converges or diverges. Explain how the test permits you to draw your conclusion.

$\Sigma^{\infty}_{n = 1}\frac{n^n}{n!}$

My Solution

$\lim_{n \to \infty}\frac{n^{n + 1}}{(n+1)n!}*\frac{n!}{n^n} = \lim_{n \to \infty}\frac{n}{n + 1} = 1$

Since $\lim_{n \to \infty}\frac{\frac{n^{n + 1}}{(n + 1)!}}{\frac{n^n}{n!}} = 1$, the ratio test fails.

Here, I have trouble thinking my solution is right because the directions make no mention of the test failing. Is my solution correct?

For problems 2 and 3 determine whether the series converges absolutely, converges conditionally, or diverges. Explain how you drew your conclusion in each case.

Problem 2.

$\Sigma^{\infty}_{n = 1}(-1)^{n}\frac{e^n}{n!}$

Problem 3.

$\Sigma^{\infty}_{n = 1}(-1)^{n}\frac{1}{\sqrt{2n + 1}}$

I don't know how to do either of these problems, help please!

Thanks in advance for all help!
(1) correction ...

$\lim_{n \to \infty}\frac{\textcolor{red}{(n+1)}^{n + 1}}{(n+1)n!}*\frac{n!}{n^n}$

(2) use the ratio test to see if $\sum \frac{e^n}{n!}$ converges ... if it does, it converges absolutely.

(3) you should already know that $\sum \frac{1}{\sqrt{2n+1}}$ diverges ... what do you know about the alternating series test?