The sum of the infinite series $\frac{1}{2}\left( \frac{1}{3} + \frac{1}{4} \right) - \frac{1}{4}\left( \frac{1}{3^{2}} + \frac{1}{4^{2}} \right) + \frac{1}{6}\left( \frac{1}{3^{3}} + \frac{1}{4^{3}} \right) - \ \ldots$ is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
$\frac{1}{2}\left( \frac{1}{3} + \frac{1}{4} ight) - \frac{1}{4}\left( \frac{1}{3^{2}} + \frac{1}{4^{2}} ight) + \frac{1}{6}\left( \frac{1}{3^{3}} + \frac{1}{4^{3}} ight) - \ldots$
$= \frac{1}{2}\left\lbrack \frac{1}{3} - \frac{1}{2}\left( \frac{1}{3^{2}} ight) + \frac{1}{3}\left( \frac{1}{3^{3}} ight) - ... ightbrack + \frac{1}{2}\left\lbrack \frac{1}{4} - \frac{1}{2}\left( \frac{1}{4^{2}} ight) + \frac{1}{3}\left( \frac{1}{4^{3}} ight) - ... ightbrack$
$= \frac{1}{2}\left\lbrack \log\left( 1 + \frac{1}{3} ight) ightbrack + \frac{1}{2}\left\lbrack \log\left( 1 + \frac{1}{4} ight) ightbrack$
$= \frac{1}{2}\left\lbrack \log{\left( \frac{4}{3} ight) 	imes \left( \frac{5}{4} ight)} ightbrack$
$= \frac{1}{2}\log\left( \frac{5}{3} ight)$

Q. The sum of the infinite series $\frac{1}{2}\left( \frac{1}{3} + \frac{1}{4} \right) - \frac{1}{4}\left( \frac{1}{3^{2}} + \frac{1}{4^{2}} \right) + \frac{1}{6}\left( \frac{1}{3^{3}} + \frac{1}{4^{3}} \right) - \ \ldots$ is equal to

Explanation

Reviewed by Dr. Abubacker

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