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pop000 · Jan 24, 2011, 02:09 PM

I need help how to solve it.

Thanks.

ebaines · Jan 24, 2011, 04:01 PM

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?

pop000 · Jan 25, 2011, 01:52 AM

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.

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

galactus · Jan 25, 2011, 04:05 AM

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

pop000 · Jan 25, 2011, 04:13 AM

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.

galactus · Jan 25, 2011, 04:37 AM

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.