# Solve: \sum_{n =0}^\infty \frac {-14} {(196n^2 - 49)} =

## Question:

Solve:

{eq}\sum_{n =0}^\infty \frac{-14}{(196n^2 - 49)} = {/eq}

## Sum of a Series:

When the term of the series is decreasing and its limit is zero, then the series will converge, And if the sigma notation form of the series is given, then we expand the terms to find the approximate sum of the series.

## Answer and Explanation:

The series can be rw-written as:

{eq}\sum _{n\:=0}^{\infty }\:\frac{-14}{\left(\left(13n\right)^2\:-\:7^2\right)}\\ =\sum _{n\:=0}^{\infty }\: \left [\frac{1}{\left(13n\:+\:7\right)}-\frac{1}{\left(13n\:-\:7\right)} \right ]\\ {/eq}

On expanding the terms, we get:

{eq}=\left (\frac{1}{7}-\frac{1}{-7} \right ) + \left ( \frac{1}{20}-\frac{1}{6} \right )+ \left ( \frac{1}{33}-\frac{1}{19} \right )+\left ( \frac{1}{46}-\frac{1}{32} \right )+ \left ( \frac{1}{59}-\frac{1}{45} \right )+ \left (\frac{1}{72} -\frac{1}{58} \right ).....\\ =0.2857-0.11 -0.02-0.0095\\ \approx 0.14 {/eq}