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1 Answer

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1 votes
We can rewrite the series as,

$\Rightarrow  \large \left ( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +..... \right ) + \left ( \frac{1}{3} + \frac{1}{9} + \frac{1}{27} +..... \right ) - \left ( \frac{1}{6} + \frac{1}{12} + \frac{1}{18} +..... \right )$

$\Rightarrow  \large \left ( \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} +..... \right ) + \left ( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} +..... \right ) - \frac{1}{6}\left ( 1 + \frac{1}{2} + \frac{1}{3} +..... \right )$

as first 2 series are are GP and value of $r < 1$, so the series converges but the later one is harmonic series,

$\Large \sum_{1}^{\infty}\frac{1}{n}$ is a harmonic series and diverges which means sum of this series is $- \infty$

So whatever may be the value of $\Rightarrow  \large \left ( \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} +..... \right ) + \left ( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} +..... \right )$ we don't need to calculate,

answer is $- \infty$

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