Wribreeminsl

2021-01-19

Find the sum of the series:
$\sum _{n=0}^{\mathrm{\infty }}\left(\left(\frac{5}{{2}^{n}}\right)-\left(\frac{1}{{3}^{n}}\right)\right)$

yunitsiL

The given series is
$\sum _{n=0}^{\mathrm{\infty }}\left(\left(\frac{5}{{2}^{n}}\right)-\left(\frac{1}{{3}^{n}}\right)\right)$
Rewrite the given series as shown below:
$\sum _{n=0}^{\mathrm{\infty }}\left(\left(\frac{5}{{2}^{n}}\right)-\left(\frac{1}{{3}^{n}}\right)\right)=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{5}{{2}^{n}}\right)-\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{{3}^{n}}\right)$
From above it can be observed that given series is the sum of two geometric series. Hence, the sum of series is obtained as,
$\sum _{n=0}^{\mathrm{\infty }}\left(\left(\frac{5}{{2}^{n}}\right)-\left(\frac{1}{{3}^{n}}\right)\right)=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{5}{{2}^{n}}\right)-\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{{3}^{n}}\right)$
$=5\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{{2}^{n}}\right)-\sum _{n=0}^{\mathrm{\infty }}\left(\frac{1}{{3}^{n}}\right)$
$=5\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)-\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)$
$=5\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)-\left(\frac{\frac{1}{3}}{\frac{2}{3}}\right)$
$=5\left(1\right)-\left(\frac{1}{2}\right)$
$=\frac{9}{2}$

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