Question

divergence/convergence testing

Answer 1

\[\sum_{n=1}^{oo}\frac{ 8 }{ n(n+2) }\]

Answer 2

you may compare it with p-series
what tests are you familiar with ?

Answer 3

im somewhat familiar with p series

Answer 4

it converges then, but now i gotta find out what it converges to

Answer 5

did you try partial fractions and see if it telescopes ?

Answer 6

did you try partial fractions and then tee

Answer 7

nope

Answer 8

so if i do partial fraction decomp on it first wouldnt it be in the form ax+b/n^2+2n

Answer 9

http://www.wolframalpha.com/input/?i=partial+fractions+8%2F%28n%28n%2B2%29%29

Answer 10

ah ok

Answer 11

\[\begin{align}\sum_{n=1}^{\infty}\frac{ 8 }{ n(n+2) }&=4\sum_{n=1}^{\infty}\left(\frac{ 1 }{ n } -\frac{1}{n+2}\right)\\~\\
&=4\sum_{n=1}^{\infty}\left(\frac{ 1 }{ n }\color{blue}{ -\frac{1}{n+1} }\right) + 4\sum_{n=1}^{\infty}\left(\color{blue}{\frac{1}{n+1}}-\frac{1}{n+2}\right)\\~\\

\end{align}\]

Answer 12

see if you can finish it off

Answer 13

how did you know to split it into 2 parts at the end there.

Answer 14

thats a good question!
i thought of splitting because we need the expression to be of form \(f(n) - f(n+1)\) for telescoping

\[\sum\limits_{n=1}^{\infty}f(n) - f(n+1) = f(1)\]

provided \(\lim\limits_{n\to\infty} f(n) = 0\)

Answer 15

\[\begin{align}\sum_{n=1}^{\infty}\frac{ 8 }{ n(n+2) }&=4\sum_{n=1}^{\infty}\left(\frac{ 1 }{ n } -\frac{1}{n+2}\right)\\~\\
&=4\sum_{n=1}^{\infty}\left(\frac{ 1 }{ n }\color{blue}{ -\frac{1}{n+1} }\right) + 4\sum_{n=1}^{\infty}\left(\color{blue}{\frac{1}{n+1}}-\frac{1}{n+2}\right)\\~\\

&=4\left(\frac{ 1 }{ 1 }\right) + 4\left(\color{blue}{\frac{1}{1+1}}\right)\\~\\
&=6
\end{align}\]

Answer 16

alrigt cool thanks

Answer 17

yw