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Question Number 74863 by mrS last updated on 02/Dec/19

Commented by abdomathmax last updated on 03/Dec/19

$$wehaveS=\sum _{n=1}^{45}\frac{1}{n^2}=\sum _{p=1}^{\left[\frac{45}{2}\right]}\frac{1}{{\left(2p\right)}^2}+\sum _{p=0}^{\left[\frac{45-1}{2}\right]}\frac{1}{{(2p+1)}^2}$$$$=\frac{1}{4}\sum _{p=1}^{22}\frac{1}{p^2}+\sum _{p=0}^{22}\frac{1}{{(2p+1)}^2}$$$$\sum _{p=1}^{22}\frac{1}{p^2}=\sum _{k=1}^{11}\frac{1}{{\left(2k\right)}^2}+\sum _{k=0}^{\left[\frac{22-1}{2}\right]}\frac{1}{{(2k+1)}^2}$$$$=\frac{1}{4}\sum _{p=1}^{11}\frac{1}{p^2}+\sum _{p=0}^{10}\frac{1}{{(2p+1)}^2}\Rightarrow$$$$S=\frac{1}{4}(\frac{1}{4}\sum _{p=1}^{11}\frac{1}{p^2}+\sum _{p=0}^{10}\frac{1}{{(2p+1)}^2})+\sum _{p=0}^{22}\frac{1}{{(2p+1)}^2}$$$$=\frac{1}{4^2}\sum _{p=1}^{11}\frac{1}{p^2}+\frac{1}{4}\sum _{p=0}^{10}\frac{1}{{(2p+1)}^2}+\sum _0^{22}\frac{1}{{(2p+1)}^2}$$$$also\sum _{p=1}^{11}\frac{1}{p^2}=\sum _{p=1}^{10}\frac{1}{4p^2}+\sum _{p=0}^5\frac{1}{{(2p+1)}^2}\Rightarrow$$$$S=\frac{1}{4^2}(\frac{1}{4}\sum _{p=1}^{10}\frac{1}{p^2}+\sum _{p=0}^5\frac{1}{{(2p+1)}^2})+\frac{1}{4}\sum _{p=0}^{10}\frac{1}{{(2p+1)}^2}$$$$+\sum _{p=0}^{22}\frac{1}{{(2p+1)}^2}$$$$S=\frac{1}{4^3}\sum _{p=1}^{10}\frac{1}{p^2}+\frac{1}{4^2}\sum _{p=0}^5\frac{1}{{(2p+1)}^2}+\frac{1}{4}\sum _{p=0}^{10}\frac{1}{{(2p+1)}^2}$$$$+\sum _{p=0}^{22}\frac{1}{{(2p+1)}^2}$$$$S=\frac{1}{4^3}\sum _{p=1}^{10}\frac{1}{p^2}+(\frac{1}{4^2}+\frac{1}{4})\sum _{p=0}^5\frac{1}{{(2p+1)}^2}+\frac{1}{4}\sum _{p=6}^{10}\frac{1}{{(2p+1)}^2}$$$$+(\frac{1}{4}+1)\sum _{p=0}^{10}\frac{1}{{(2p+1)}^2}+\sum _{p=11}^{22}\frac{1}{{(2p+1)}^2}$$$$\sum _{p=6}^{10}\frac{1}{{(2p+1)}^2}{=}_{p-6=k}\sum _{k=0}^4\frac{1}{{(2(6+k)+1)}^2}$$$$=\sum _{k=0}^4\frac{1}{{(2k+13)}^2}$$$$\sum _{p=11}^{22}\frac{1}{{(2p+1)}^2}{=}_{p-11=k}\sum _{k=0}^{11}\frac{1}{{(2k+23)}^2}$$$$and\sum _{p=1}^{10}\frac{1}{p^2}canalsobesimplifiedsothe$$$$valueofSisp.pknown....$$

Answered by MJS last updated on 02/Dec/19

$$\mathrm{all}\mathrm{you}\mathrm{can}\mathrm{do}\mathrm{is}\mathrm{calculate}\mathrm{it}$$

Commented by tw000001 last updated on 03/Dec/19

$$\mathrm{I}\mathrm{know}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{6}\approx 1.644,$$$$\mathrm{then}\underset{n=1}{\sum ^{45}}\frac{1}{n^2}\approx 1.622.$$

Answered by tw000001 last updated on 03/Dec/19

$$\underset{n=1}{\sum ^{45}}\frac{1}{n^2}=\frac{{\pi }^2}{6}-\frac{1}{45}\approx 1.622$$

Commented by MJS last updated on 03/Dec/19

$$\underset{n=1}{\sum ^{45}}\frac{1}{n^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}-\frac{1}{45}$$$$\frac{1}{45}+\underset{n=1}{\sum ^{45}}\frac{1}{n^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}$$$$(k\in {\mathbb{N}}^{★}\Rightarrow \underset{n=1}{\sum ^k}\frac{1}{n^2}\in \mathbb{Q})\wedge \frac{1}{45}\in \mathbb{Q}\Rightarrow$$$$\Rightarrow (\frac{1}{45}+\underset{n=1}{\sum ^{45}}\frac{1}{n^2})\in \mathbb{Q}\Rightarrow \frac{{\pi }^2}{6}\in \mathbb{Q}\mathrm{which}\mathrm{is}\mathrm{wrong}$$$$\Rightarrow \underset{n=1}{\sum ^{45}}\frac{1}{n^2}\ne \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}-\frac{1}{45}$$

Commented by tw000001 last updated on 03/Dec/19

$$\mathrm{OK},\mathrm{I}\mathrm{know}\mathrm{the}\mathrm{answer}\mathrm{is}1.622.$$$$\mathrm{But}\mathrm{is}\mathrm{there}\mathrm{any}\mathrm{possible}\mathrm{solution}\mathrm{that}\mathrm{is}$$$$\mathrm{approximately}\mathrm{on}1.622?$$

Commented by MJS last updated on 03/Dec/19

$$\mathrm{the}\mathrm{exact}\mathrm{value}\mathrm{is}\frac{m}{n}\mathrm{with}m,n>{10}^{37}$$$$\mathrm{you}\mathrm{can}\mathrm{use}\mathrm{a}\mathrm{calculator}\mathrm{to}\mathrm{get}\mathrm{an}\mathrm{approximation}$$$$\mathrm{but}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{unique}\mathrm{approximation}$$$$\mathrm{I}\mathrm{get}$$$$\underset{n=1}{\sum ^{45}}\frac{1}{n^2}\approx 1.6229569293973$$$$\mathrm{the}\mathrm{exact}\mathrm{value}\mathrm{is}$$$$\frac{2i0571823268450072862674893950786869803}{12675520154492970709544386574878080000}$$

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