Find the sum of the series. sum_{n=1}^inftyfrac{2}{(n+8)(n+6)}

Braxton Pugh 2020-11-24 Answered
Find the sum of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{2}{\left(n+8\right)\left(n+6\right)}$
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aprovard
The series is:
$\sum _{n=1}^{\mathrm{\infty }}\frac{2}{\left(n+8\right)\left(n+6\right)}$
Simplify the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{2}{\left(n+8\right)\left(n+6\right)}=\sum _{n=1}^{\mathrm{\infty }}\frac{8-6}{\left(n+8\right)\left(n+6\right)}$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{\left(n+8\right)-\left(n+6\right)}{\left(n+8\right)\left(n+6\right)}$
$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n+6\right)}-\frac{1}{\left(n+8\right)}$
Expand the summation.
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n+6\right)}-\frac{1}{\left(n+8\right)}=\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}...\right)-\left(\frac{1}{9}+\frac{1}{10}+...\right)$
$=\frac{1}{7}+\frac{1}{8}$
$=\frac{15}{56}$
Jeffrey Jordon