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How can I find if the following converges? I tried using the comparaison theorem but I don't know to what compare it? And I tried finding the Sum of the series but I can wrong results each time.. $$\sum_{x=1}^\infty \frac{1}{(2x-1)(2x+1)}$$

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    As far as convergence, the expand the denominator. – 2017-01-13
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    I found by expanding that I weill get the sums of Sum A + Sum B/(k-1/2) + Sum C/(k+1/2). for the first 2 sums they are clearly divergent but how can I show the serie C/(k+1/2) will diverge? – 2017-01-13

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For a limit comparison test:

$$\frac1{(2x-1)(2x+1)}\sim\frac1{4x^2}$$

and compare to the p-series.

Partial fraction decomposition gives us

$$\frac1{(2x-1)(2x+1)}=\frac12\left(\frac1{2x-1}-\frac1{2x+1}\right)$$

So that we get a telescoping series

$$\sum_{x=1}^n\frac12\left(\frac1{2x-1}-\frac1{2x+1}\right)=\frac12-\frac1{2(2n+1)}$$

$$\implies\sum_{x=1}^\infty\frac1{(2x-1)(2x+1)}=\frac12$$

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Hint: $\require{cancel}\frac1{(2x-1)(2x+1)}$ can be written as $\frac12(\frac1{2x-1}-\frac1{2x+1})$. The first few partial sums are then $$S_1=\tfrac12(\tfrac11-\tfrac13) = \frac12(1-\tfrac13)$$ $$S_2=\tfrac12(\tfrac11-\cancel{\tfrac13}) + \tfrac12(\cancel{\tfrac13}-\tfrac15) = \tfrac12(1-\tfrac15)$$ $$S_3=\tfrac12(\tfrac11-\cancel{\tfrac13}) + \tfrac12(\cancel{\tfrac13}-\cancel{\tfrac15}) + \tfrac12(\cancel{\tfrac15}-\tfrac17) = \tfrac12(1-\tfrac17)$$

You should be able to write a formula for $S_n$ and then find the desired limit $\lim_{n\to\infty}S_n$.