telescoping series

[missyc8]

determine whether the series converges, and if so find the sum

summation of k=1 to infinity (1/(k+2)(k+3)

do i use parts and if yes, where do i go from there ..

 

[akoaysigod]

You integrate using a technique I don't know the name of. That would equal

A/(k+2) + B/(k+3) = int(1/(k+2)(k+3) then multiply both sides by (k+2)(k+3) then solve for the different A and Bs.
Ak + Bk = 0 2A + 3B = 1
I would check that last part though since I didn't write it down its probably wrong. Then you put whatever you get for A and B back into the left hand side and integrate them separately; you'll get two something like 1/c[ln(k+2)+ln(k+3)].

You might be able to integrate by parts but I don't know.

Then you take the limit as t approaches infinity of whatever you got for the integral from 1 to t.

Someone else should probably check that though. I don't know what a telescoping series is but I am integrating that correctly.

 

[BigGlenntheHeavy]

\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{(k+2)(k+3)} \ = \sum_{k=1}^{\infty}\bigg[\frac{1}{k+2}-\frac{1}{k+3}\bigg]\)

\(\displaystyle = \ \bigg(\frac{1}{3}-\frac{1}{4}\bigg)+\bigg(\frac{1}{4}-\frac{1}{5}\bigg)+\bigg(\frac{1}{5}-\frac{1}{k+3}\bigg)+...\)

\(\displaystyle = \ \lim_{k\to\infty} \ \bigg[\frac{1}{3}-\frac{1}{k+3}\bigg] \ = \ \frac{1}{3}.\)

\(\displaystyle Hence, \ summation \ converges \ to \ \frac{1}{3}.\)