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



{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}

Learn more about this topic:

Using Sigma Notation for the Sum of a Series

from Algebra II Textbook

Chapter 21 / Lesson 13

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