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Question Number 24549 by math solver last updated on 20/Nov/17

Commented by math solver last updated on 20/Nov/17

$$\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{ratio}\mathrm{is}??$$

Commented by prakash jain last updated on 20/Nov/17

$$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...=\mathrm{S}$$$$(1+\frac{1}{3^2}+\frac{1}{5^2}+..)+\frac{1}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+...)=\mathrm{S}$$$$(1+\frac{1}{3^2}+\frac{1}{5^2}+..)+\frac{\mathrm{S}}{4}=\mathrm{S}$$$$(1+\frac{1}{3^2}+\frac{1}{5^2}+..)=\frac{3\mathrm{S}}{4}$$$$\mathrm{Given}\mathrm{ratio}$$$$\frac{\mathrm{S}}{\frac{3\mathrm{S}}{4}-\frac{\mathrm{S}}{4}}=2$$

Commented by prakash jain last updated on 20/Nov/17

$$\mathrm{Terms}\mathrm{can}\mathrm{be}\mathrm{rearranged}\mathrm{for}$$$$\mathrm{absolutely}\mathrm{convergent}\mathrm{series}.$$

Commented by math solver last updated on 20/Nov/17

$$\mathrm{god}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by maxmathsup by imad last updated on 24/May/19

$$letq=\frac{1+\frac{1}{2^2}+\frac{1}{3^2}+...}{1-\frac{1}{2^2}+\frac{1}{3^2}+..}\Rightarrow q=\frac{\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}}{\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}}$$$$wehave\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\left(2n\right)}^2}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=-\frac{1}{4}\frac{{\pi }^2}{6}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=\frac{3}{4}.\frac{{\pi }^2}{6}=\frac{{\pi }^2}{8}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}=\frac{{\pi }^2}{8}-\frac{{\pi }^2}{24}=\frac{{\pi }^2}{12}\Rightarrow q=\frac{\frac{{\pi }^2}{6}}{\frac{{\pi }^2}{12}}\Rightarrow q=2.$$

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