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

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.