1. ## Series Convergences....

This is a simple series, it was listed in the review section of my book, they state for this problem determine if the series converges or diverges by using the integral test, however, I feel limiting myself to one method of showing convergence will not help me on the test, so I need someone to check my different ways of showing this converges

The initial problem $\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n+2)^2}$

Using the integral test

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n+2)^2} = \lim_{b\to\infty}\int_1^b \frac{1}{(x+2)^2} dx$

Using some simple substitution to find the integrand of
$\displaystyle \lim_{b\to\infty}- \frac{1}{(x+2)} |_1^b$

Then

$\displaystyle \lim_{b\to\infty}- \frac{1}{(b+2)} + \lim_{b\to\infty}\frac{1}{(3)}=\frac{1}{3}$

Since the limit is finite the series converges.

Using a direct comparison

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n+2)^2}\le\sum_{n=1}^ {\infty}\frac{1}{n^2}$

We know by the p-series test for $\displaystyle p>1$ converges, thus the series converges

We can also take the limit of the series itself and apply L'Hopital's rule and see the limit is 0 but this doesn't mean much, because it could also diverge or converge

2. definitely, the best and faster way, it's the comparison test.

3. It seems so in the cases your can use it. I am sure you can use it in any case but my skills are not that great, I work with what I can and improve with I have little by little. It seems in most scenarios the integral test is recommend with fraction with a higher power in the denomiator