Some special manipulations involving finite sums. How to solve this sum?

$\displaystyle{\sum_{k=1}^{n}}\frac{1}{4k^2 - 1}$

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    $\begingroup$ Hint 1: the denominator can be factorised. Hint 2: Partial fractions. Hint 3: It telescopes. $\endgroup$ – ShreevatsaR Oct 16 '11 at 17:09

Hint: Note that $$\frac{1}{4k^2-1}=\frac{\frac{1}{2}}{2k-1}-\frac{\frac{1}{2}}{2k+1}.$$


The following problem below is for an infinite series, but if you can solve it you may be able to solve your problem above.

  1. Let $a_n := b_{n} - b_{n+1}$, for some other sequence $b_n$. Prove that the series $\sum_{n =0}^{\infty} a_n$ converges iff the sequence $b_n$ does.

  2. In the case that the sum converges, what is its sum?

  3. Use (1) and (2) to show that $\sum_{k=0}^{\infty} \frac{1}{4k^2 - 1}$ converges and find its sum.


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