Sum and convergence of Series

**Benno_11** · May 23rd 2009, 05:57 PM

Hey,

I need some help finding the sum of this and showing it converges. To show a series is convergent do you show its limit approaches zero? So for a divergent series you would show it doesn't approach zero.

$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}$

Thanks

**Sampras** · May 23rd 2009, 06:00 PM

Originally Posted by Benno_11

Hey,

I need some help finding the sum of this and showing it converges. To show a series is convergent do you show its limit approaches zero? So for a divergent series you would show it doesn't approach zero.

$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}$

Thanks

No, the test is inconclusive when the limit of the summand is zero (eg. the series can converge or diverge). Try the ratio test.

**Media_Man** · May 23rd 2009, 06:09 PM

Use partial fractions.

$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ = $\sum_{n=1}^\infty \frac{A}{n^2}+ \frac{B}{(n+1)^2}$ for some integers A,B

$\sum_{n=1}^\infty \frac{A}{n^2}+ \frac{B}{(n+1)^2}$ = $\sum_{n=1}^\infty \frac{A(n+1)^2+Bn^2}{n^2(n+1)^2}$ = $\sum_{n=1}^\infty \frac{(A+B)n^2+2An+1}{n^2(n+1)^2}$

So A+B=0 and 2A=2, therefore A=1 and B=-1, so

$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ = $\sum_{n=1}^\infty \frac{1}{n^2}- \frac{1}{(n+1)^2}$ = $\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16} +...$

Notice all but the last terms cancel each other, so...

$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ = $\lim_{L\rightarrow\infty} 1-\frac{1}{L^2}=1$

QED

**skeeter** · May 23rd 2009, 06:18 PM

Originally Posted by Benno_11

Hey,

I need some help finding the sum of this and showing it converges. To show a series is convergent do you show its limit approaches zero? no

So for a divergent series you would show it doesn't approach zero.

if the nth term does not approach zero, the series diverges ... but note that there are series whose nth terms do approach zero and diverge.

$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}$

Thanks

the series is a telescoping series. using the method of partial fractions ...

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

list out a few terms and a pattern emerges, note that the inner terms cancel ...

$\left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \left(\frac{1}{9} - \frac{1}{16}\right) + ... + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) + ... = \lim_{n \to \infty} 1 - \frac{1}{(n+1)^2} = 1$

**Krizalid** · May 23rd 2009, 06:38 PM

convergence is pretty easy to prove since for each $n\ge1$ is $\frac{2n+1}{n^{2}(n+1)^{2}}< \frac{3n}{n^{4}}=\frac{3}{n^{3}},$ thus $\sum\limits_{n=1}^{\infty }{\frac{2n+1}{n^{2}(n+1)^{2}}}< \sum\limits_{n=1}^{\infty }{\frac{3}{n^{3}}}<\infty .$ (Direct comparison test with a $p-$ series.)

now, to split the original fraction into two ratios just use algebra!

$\frac{2n+1}{n^{2}(n+1)^{2}}=\frac{\left( n^{2}+2n+1 \right)-n^{2}}{n^{2}(n+1)^{2}}=\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}},$ which is clearly a telescoping series as skeeter stated above.

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