The sum of the series $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \ldots \ldots \ldots \ldots .$. up to $\infty$ is equal to

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Aicharya · Accepted Answer

$\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots\infty$ Let $T_n=\frac{1}{n(n+1)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)$ $\mathrm{S}=\mathrm{T}_1-\mathrm{T}_2+\mathrm{T}_3-\mathrm{T}_4+\mathrm{T}_5 \ldots \ldots \ldots \ldots \ldots$ $=\left(\frac{1}{1}-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{5}\right) \ldots \ldots \ldots$. $=1-2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5} \ldots \ldots \ldots \ldots \infty\right]$ $=1-2[-\log (1+1)+1]=2 \log 2-1=\log \left(\frac{4}{e}\right)$

The sum of the series $\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \ldots \ldots \ldots \ldots …

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