# Find the sum of a series

• Oct 19th 2013, 10:39 PM
Memorystack
Find the sum of a series
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)
Attachment 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
• Oct 19th 2013, 11:35 PM
SlipEternal
Re: Find the sum of a series
Let's look at partial sums: $\displaystyle S_k = \sum_{n=1}^k \dfrac{1}{4n^2-1}$. We have

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

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

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

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

So, $\displaystyle \sum_{n=1}^\infty \dfrac{1}{4n^2-1} = \lim_{k \to \infty} \dfrac{k}{2k+1} = \dfrac{1}{2}$.
• Oct 20th 2013, 03:09 AM
Prove It
Re: Find the sum of a series
Quote:

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)
Attachment 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 \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.