Find the sum of a series

**Memorystack** · Oct 19th 2013, 10:39 PM

Hello there,

I'm a litttle bit stuck on a math question here

I need to find the Sum of the serie 1/((4*i^2)-1)

Find the sum of a series-sum-problem.png

*made a mistake in the picture the i element is suppose to be a n sorry

I can't seem to understand how to even start

thank you guys

**SlipEternal** · Oct 19th 2013, 11:35 PM

Let's look at partial sums: $S_k = \sum_{n=1}^k \dfrac{1}{4n^2-1}$ . We have

$S_1 = \dfrac{1}{3}$

$S_2 = \dfrac{1}{3} + \dfrac{1}{15} = \dfrac{2}{5}$

$S_3 = \dfrac{2}{5} + \dfrac{1}{35} = \dfrac{3}{7}$

In general, the pattern appears to be $S_k = \dfrac{k}{2k+1}$ . You can show inductively that this is the case.

So, $\sum_{n=1}^\infty \dfrac{1}{4n^2-1} = \lim_{k \to \infty} \dfrac{k}{2k+1} = \dfrac{1}{2}$ .

**Prove It** · Oct 20th 2013, 03:09 AM

Originally Posted by Memorystack

Hello there,

I'm a litttle bit stuck on a math question here

I need to find the Sum of the serie 1/((4*i^2)-1)

Click image for larger version.

Name: Sum problem.png
Views: 11
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ID: 29527

*made a mistake in the picture the i element is suppose to be a n sorry

I can't seem to understand how to even start

thank you guys

Notice that $\displaystyle \begin{align*} \frac{1}{4n^2 - 1} = \frac{1}{(2n - 1)(2n + 1)} \end{align*}$

Split it up into partial fractions and you should find that this is a telescopic series.

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