# Series Convergences....

• Nov 27th 2009, 05:17 PM
RockHard
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
• Nov 27th 2009, 06:16 PM
Krizalid
definitely, the best and faster way, it's the comparison test.
• Nov 27th 2009, 06:42 PM
RockHard
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