Ask Me Help Desk - Prove that the series are Convergence

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Prove that the series are Convergence

I need help how to solve it.

http://p1cture.me/images/08524442504815639122.jpg
Thanks.

Combine the two fractions into a single fraction, and notice how the denominator now has a factor of n^2. Can you take it from here?

well after I Multiply the numerator of the other denominator of each Fracture I got the Common denominator that you can see in the picture.
http://p1cture.me/images/16081112965142380290.jpg

and yes there n^2 so this is the answer?

You did not state the summation limits.

Assuming they are:

$\sum_{n=1}^{\infty}\left(\frac{1}{n+2}-\frac{1}{n+3}\right)$

Notice this is a 'telescoping' sum.

$\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+.....$

See how everything cancels out except for the first term... 1/3

The sum is 1/3

Comment on galactus's post

Hi. Thanks for answer but I need to use the condensation test.
Can you show me any way so solve this with the condensation test?
Thank you.

Cauchy's condensation test says that the series is convergent if

$\sum_{n=0}^{\infty}2^{n}f(2^{n})$ is convergent.

Using this, leads to

$\sum_{n=0}^{\infty}\frac{2^{n}}{(2^{n}+2)(2^{n}+3) }$

Now, one way may be to use the ratio test. This leads to:

$\frac{(2^{n}+2)(2^{n}+3)}{(2^{n}+1)(2^{n+1}+3)}$

Taking the limit $n\to \infty$ , leads to a limit of 1/2.

Since the limit is <1, it is convergent.

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