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Question Number 99889 by Dwaipayan Shikari last updated on 23/Jun/20

$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+.......\mathrm{\infty }\left\{\mathrm{Find}\mathrm{the}\mathrm{sum}\right\}$$

Answered by smridha last updated on 23/Jun/20

$$\mathit{t}\mathit{h}\mathit{i}\mathit{s}\mathit{i}\mathit{s}\mathit{c}\mathit{a}\mathit{l}\mathit{l}\mathit{e}\mathit{d}\mathit{H}a\mathit{r}\mathit{m}\mathit{o}\mathit{n}\mathit{i}\mathit{c}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}$$$$\mathit{a}\mathit{n}\mathit{d}\mathit{H}a\mathit{r}\mathit{m}\mathit{o}\mathit{n}\mathit{i}\mathit{c}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}\mathit{a}\mathit{l}\mathit{w}\mathit{a}\mathit{y}\mathit{s}$$$$\mathit{d}\mathit{i}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{s}\mathit{o}\mathit{i}\mathit{t}\mathit{h}\mathit{a}\mathit{s}\mathit{n}\mathit{o}\mathit{f}\mathit{i}\mathit{n}\mathit{i}\mathit{t}\mathit{e}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}$$$${\mathit{H}}_{\mathit{n}}=\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathit{n}}=\mathrm{\infty }$$$$\mathit{i}\mathit{f}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}\mathit{i}\mathit{s}\mathit{l}\mathit{i}\mathit{k}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}:$$$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...\mathrm{\infty }=\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n}{\left(1\right)}^{\mathit{n}}$$$$=\mathit{l}\mathit{n}\left(2\right)\approx 0.693$$$$$$

Commented by Dwaipayan Shikari last updated on 23/Jun/20

$$Ohhitisalogarithmicseries.Thankingyou$$

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