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Question Number 96667 by Ar Brandon last updated on 03/Jun/20

$$\mathcal{P}\mathrm{rove}\mathrm{that}\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{k}}^2}=\frac{{\pi }^2}{6}$$

Answered by Sourav mridha last updated on 03/Jun/20

$$\underset{\mathit{k}=1}{\sum ^{\mathrm{\infty }}}{\mathit{k}}^{-2}=\mathit{\zeta }\left(2\right)$$$$\mathit{f}\mathrm{ro}\mathit{m}\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathit{l}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{o}\mathit{f}\mathit{R}\mathit{i}\mathrm{e}\mathit{m}\mathit{a}\mathit{n}\mathit{n}$$$$\mathit{z}\mathit{e}\mathit{t}\mathit{a}{\mathit{f}}^{\mathit{n}}\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathit{B}\mathit{u}\mathit{r}\mathit{n}\mathit{o}\mathit{u}\mathit{l}\mathit{l}\mathit{i}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}$$$$\left(\mathit{e}\mathit{v}\mathit{e}\mathit{n}\right).$$$${\mathit{B}}_{2\mathit{n}}={(-1)}^{\mathit{n}-1}\frac{2\left(2\mathit{n}\right)!}{{\left(2\mathit{\pi }\right)}^{2\mathit{n}}}.\mathit{\zeta }\left(2\mathit{n}\right).$$$$\mathit{p}\mathit{u}\mathit{t}..\mathit{n}=1$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{n},{\mathit{B}}_2=\frac{2.2!}{2.2.{\mathit{\pi }}^2}.\mathit{\zeta }\left(2\right)$$$$\mathit{w}\mathit{e}\mathit{k}\mathit{n}\mathit{o}\mathit{w}{\mathit{B}}_2=\frac{1}{6}.\mathit{n}\mathit{o}\mathit{w}\mathit{p}\mathit{u}\mathit{t}\mathit{t}\mathit{i}\mathit{n}\mathit{g}\mathit{i}\mathit{t}$$$$\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}..\mathit{\zeta }\left(2\right)=\frac{{\mathit{\pi }}^2}{6}...$$

Commented by I want to learn more last updated on 05/Jun/20

$$\mathrm{sir}\mathrm{what}\mathrm{of}\zeta (2\mathbf{n}+1)\mathbf{for}\mathbf{odd}$$

Commented by Sourav mridha last updated on 05/Jun/20

$$\mathit{T}\mathrm{his}\mathit{i}\mathit{s}\mathit{n}\mathit{o}\mathit{t}\mathit{j}\mathit{u}\mathit{s}\mathit{t}\mathit{i}\mathit{f}\mathit{i}\mathit{e}\mathit{d}\mathit{b}\mathit{y}\mathit{t}\mathit{h}\mathit{i}\mathit{s}$$$$\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}\mathit{b}\mathit{e}\mathit{c}\mathit{a}\mathit{u}\mathit{s}\mathit{e}:$$$${\mathit{B}}_{2\mathit{n}+1}=0\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathit{n}\in \mathbb{N}$$$$\mathit{o}\mathit{n}\mathit{l}\mathit{y}\mathit{o}\mathit{d}\mathit{d}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}{\mathit{B}}_1=-\frac{1}{2}\mathit{e}\mathit{x}\mathit{i}\mathit{s}\mathit{t}.$$$$\mathit{g}\mathit{e}\mathit{n}\mathit{e}\mathit{r}\mathit{e}\mathit{l}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{o}\mathit{f}{\mathit{B}}_{\mathit{n}}:$$$${\mathit{B}}_{\mathit{n}}=-\frac{\mathit{n}!}{{\left(2\mathit{\pi }\mathit{i}\right)}^{\mathit{n}}}\left[\underset{\mathrm{m}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{{\mathit{m}}^{\mathit{n}}}+\frac{1}{{(-\mathit{m})}^{\mathit{n}}})\right]$$$$$$$$$$

Commented by I want to learn more last updated on 06/Jun/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by Rio Michael last updated on 03/Jun/20

$${\mathrm{i}}^{\prime }\mathrm{ve}\mathrm{always}\mathcal{W}\mathrm{anted}\mathrm{to}\mathcal{G}\mathrm{et}\mathrm{this}\mathrm{proof}..\mathrm{and}{\mathrm{here}}^{\prime }\mathrm{s}\mathrm{what}\mathrm{i}\mathrm{fell}\mathrm{on}.$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}=\frac{{\pi }^2}{6}$$$$\mathrm{first}\mathrm{let}s=\arcsin x$$$$\mathrm{hence}{\stackrel{\frac{\pi }{2}}{\int }}_{0}sds=\frac{{\pi }^2}{8}$$$$\mathrm{also}{\int }_0^1\frac{\arcsin x}{\sqrt{1-x^2}}dx=\frac{{\pi }^2}{8}$$$$\mathrm{since}\mathrm{arc}\sin x={\int }_0^1\frac{1}{\sqrt{1-x^2}}dx=x+\frac{x^3}{2\times 3}+\frac{1\times 3x^5}{2\times 5\times 5}+...$$$$\mathrm{substituting}\mathrm{we}\mathrm{get}:{\int }_0^1(\frac{dx}{\sqrt{1-x^2}}\int \frac{dx}{\sqrt{1-x^2}})={\int }_0^1[x+\frac{x^3}{6}\left(\frac{dx}{\sqrt{1-x^2}}\right)+...]$$$$\mathrm{realise}\mathrm{that}:{\int }_0^1\frac{x^{2n+1}}{\sqrt{1-x^2}}dx=\frac{2n!!}{(2n+1)!!}.\mathrm{since}\mathrm{we}\mathrm{get}\mathrm{all}\mathrm{odd}\mathrm{powers},$$$$\frac{{\pi }^2}{8}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(2n-1)}^2}$$$$\mathrm{if}\mathrm{we}\mathrm{define}x=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}$$$$\mathrm{dividing}\mathrm{by}4,\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{equation}:x=\frac{{\pi }^2}{6}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}$$$$\mathrm{please}\mathrm{guys}\mathrm{check}.$$

Commented by Ar Brandon last updated on 03/Jun/20

I'm lost.��

Commented by I want to learn more last updated on 06/Jun/20

$$\mathrm{thanks}\mathrm{sir}$$

Commented by I want to learn more last updated on 06/Jun/20

$$\mathrm{sir},\mathrm{will}\mathrm{this}\mathrm{method}\mathrm{get}:\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\mathrm{k}}^3}$$

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