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Question Number 32755 by 7991 last updated on 01/Apr/18

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=....???$$

Commented by Rio Mike last updated on 01/Apr/18

$$\mathrm{solution}$$$$\mathrm{n}=1\Rightarrow \frac{1}{1^2}$$$$\mathrm{n}=2\Rightarrow \frac{1}{2^2}$$$$\mathrm{n}=3\Rightarrow \frac{1}{2^3}$$$$\mathrm{the}\mathrm{GP}\mathrm{is}1,\frac{1}{4},\frac{1}{8},...$$$${\mathrm{S}}_{\mathrm{\infty }}=\frac{\mathrm{a}}{1-\mathrm{r}}$$$$\Rightarrow \frac{1}{1-\frac{1}{4}}$$$$=1\times \frac{4}{3}$$$$=\frac{4}{3}$$

Commented by abdo imad last updated on 01/Apr/18

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\xi \left(2\right)=\frac{{\pi }^2}{6}with\xi \left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^x}withx>1$$$$youwillfindtheproofinthisplatform.$$

Commented by abdo imad last updated on 01/Apr/18

$$youranswerisnotcorrectRiosir....$$

Commented by JDamian last updated on 05/Apr/18

$${Isn}^{\prime }tthisknownas{Basel}^{\prime }sproblem?$$

Commented by Rasheed.Sindhi last updated on 05/Apr/18

$$1,\frac{1}{4},\frac{1}{8},...\mathrm{is}\mathrm{not}\mathrm{GP}$$$$\because \frac{1}{4}\mathbf{÷}1\ne \frac{1}{8}\mathbf{÷}\frac{1}{4}$$$$\frac{1}{4}\ne \frac{1}{2}$$

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