I am having trouble with a couple of convergence/divergence tests for series. So far in class we've talked about the following:

- nth-term test
- geometric series test
- p-series test
- integral test
- comparison test
- ratio test
- root test

I've been working on harder problems but I am stuck. Could you point me towards the right direction and help me see the bigger picture?

$\displaystyle \sum_{n=1}^{\infty }\frac{n+1}{n^3+1}$

$\displaystyle f(x)=\frac{x+1}{x^3+1}$

$\displaystyle \frac{d}{dx}(x+1)< \frac{d}{dx}(x^3+1) \therefore f(x) \textup{ is decreasing.}$

Since the denominator grows faster than the numerator, then that implies that it diverges. According to Wolfram, the ratio and root test would be inconclusive but it would converge with the comparison test. So does it diverge or converge? What would I compare it to with the comparison test?

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

$\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow \infty}\frac{(n+1)n!}{(n+1)^{n+1}}=\lim_{n\rightar row \infty}(\frac{n}{n+1})^n$

What do I do after that? According to Wolfram the $\displaystyle \lim_{n\rightarrow \infty}(\frac{n}{n+1})^n=\frac{1}{e}<1 \therefore $ it converges. Why is it equal $\displaystyle \frac{1}{e}$?

$\displaystyle \sum_{n=1}^{\infty }3^{\frac{1}{n}}$

I have no idea how to even start this... Any suggestions?

$\displaystyle \sum_{n=1}^{\infty }\frac{1}{ln(n)^{n}}$

$\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow \infty}\frac{ln(n)^{n}}{ln(n+1)^{n+1}}=?$

How do I even begin to break that down so things start cancelling?

$\displaystyle \frac{1}{2}+\frac{1}{2}\cdot \frac{4}{4}+\frac{1}{2}\cdot \frac{4}{4}\cdot \frac{7}{6}+\cdots +\frac{1\cdot 4\cdot 7\cdots \cdots (3n-2)}{2\cdot 4\cdot 6\cdots \cdots (2n)}+\cdots $

I have no idea how to begin this either!

I know I have a lot of questions but I would greatly appreciate any help you give me. I think I'm missing a key part. Thank you so much!