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The sum of the series \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\ldots . \infty\) is
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Verified Answer
The correct answer is:
\(2 \log _{\mathrm{e}} 2-1\)
\(\begin{aligned}\text {Hints: }
& s=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \infty \\
= & \left(\frac{1}{1}-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right) \ldots . \\
= & \frac{1}{1}-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5} \ldots \\
= & \frac{1}{1}-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}+\frac{2}{5}-\ldots \\
= & 2\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} \ldots .\right]-1=2 \log 2-1
\end{aligned}\)
& s=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \infty \\
= & \left(\frac{1}{1}-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right) \ldots . \\
= & \frac{1}{1}-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5} \ldots \\
= & \frac{1}{1}-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}+\frac{2}{5}-\ldots \\
= & 2\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} \ldots .\right]-1=2 \log 2-1
\end{aligned}\)
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