# Thread: Two Infinite Series Problems

1. ## Two Infinite Series Problems

I have two problems that I can't seem to get.

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

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

My Solution

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

The problem here is, I know the ratio test isn't supposed to fail for this problem, lol. Where am I going wrong?

Problem 2.
Use the limit comparison test to determine whether the series converges or diverges. Explain how the test permits you to draw your conclusion.

$\displaystyle \Sigma^{\infty}_{n = 1}\frac{n^2}{\sqrt{2n^5 + 1}}$

My Solution

$\displaystyle = \Sigma^{\infty}_{n = 1}\frac{n^2}{\sqrt{2}n^{\frac{5}{2}} + 1}$, Compare to $\displaystyle \frac{1}{n^{\frac{1}{2}}}$, conjecture to diverge.

$\displaystyle \lim_{n \to \infty}\Sigma^{\infty}_{n = 1}\frac{n^2}{\sqrt{2}n^{\frac{5}{2}} + 1} * \frac{n^{\frac{1}{2}}}{1} = \lim_{n \to \infty}\frac{n^{\frac{5}{2}}}{\sqrt{2}n^{\frac{5}{ 2}} + 1} = \frac{1}{\sqrt{2}}$

Since $\displaystyle \lim_{n \to \infty}\frac{\frac{n^2}{\sqrt{2n^5 + 1}}}{\frac{1}{n^{\frac{1}{2}}}} = \frac{1}{\sqrt{2}}$ and $\displaystyle 0 < \frac{1}{\sqrt{2}} < \infty$, and $\displaystyle \frac{1}{n^{\frac{1}{2}}}$ diverges, $\displaystyle \frac{n^2}{\sqrt{2n^5 + 1}}$ diverges.

I was told that my algebra in this problem was incorrect, but I don't know why. Can someone show me where I went wrong, please?

Thanks in advance for any help.

2. Will answer second in a minute.

For first one your final step is wrong...

$\displaystyle \frac{(n+1)^n}{n^n} = \bigg{(} \frac{n+1}{n} \bigg{)}^n$
= $\displaystyle \bigg{(} 1 + \frac{1}{n} \bigg{)}^n = e$ (exponential function)

3. Originally Posted by mturner07

Problem 2.

My Solution

$\displaystyle = \Sigma^{\infty}_{n = 1}\frac{n^2}{\sqrt{2}n^{\frac{5}{2}} + 1}$,
This part is wrong. $\displaystyle \sqrt{2n^2 + 1} \neq \sqrt{2}n^{5/2} + 1$

You could write...
$\displaystyle \sum^{\infty}_{n = 1}\frac{n^2}{\sqrt{2}n^{\frac{5}{2}} + 1} \leq = \sum^{\infty}_{n = 1}\frac{n^2}{\sqrt{2}n^{\frac{5}{2}}}$