Question Number 132715 by Dwaipayan Shikari last updated on 16/Feb/21
$$\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{3}} }+… \\ $$
Answered by Olaf last updated on 16/Feb/21
$$\mathrm{Let}\:\left({a}_{{n}} \right)_{{n}\in\mathbb{N}} \:=\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{4},\:\mathrm{5},\:\mathrm{7},\:\mathrm{8}…\right\} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{6}{n}+\mathrm{3}+\left(−\mathrm{1}\right)^{{n}} \right] \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{a}_{{n}} ^{\mathrm{3}} } \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{4}^{\mathrm{3}} }{\left[\mathrm{6}{n}+\mathrm{3}+\left(−\mathrm{1}\right)^{{n}} \right]^{\mathrm{3}} } \\ $$$$ \\ $$$$\mathrm{S}\:=\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{p}} \mathrm{4}^{\mathrm{3}} }{\left[\mathrm{6}\left(\mathrm{2}{p}\right)+\mathrm{3}+\left(−\mathrm{1}\right)^{\mathrm{2}{p}} \right]^{\mathrm{3}} } \\ $$$$+\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{p}+\mathrm{1}} \mathrm{4}^{\mathrm{3}} }{\left[\mathrm{6}\left(\mathrm{2}{p}+\mathrm{1}\right)+\mathrm{3}+\left(−\mathrm{1}\right)^{\mathrm{2}{p}+\mathrm{1}} \right]^{\mathrm{3}} } \\ $$$$ \\ $$$$\mathrm{S}\:=\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{\mathrm{3}} }{\left(\mathrm{12}{p}+\mathrm{4}\right)^{\mathrm{3}} }−\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{\mathrm{3}} }{\left(\mathrm{12}{p}+\mathrm{8}\right)^{\mathrm{3}} } \\ $$$$\mathrm{S}\:=\:\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{p}+\mathrm{1}\right)^{\mathrm{3}} }−\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{p}+\mathrm{2}\right)^{\mathrm{3}} } \\ $$$$\mathrm{S}\:=\:−\frac{\Psi\left(\mathrm{2},\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{54}}+\frac{\Psi\left(\mathrm{2},\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{54}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{4}\pi^{\mathrm{3}} \sqrt{\mathrm{3}}}{\mathrm{243}} \\ $$$$\mathrm{S}\:\approx\:\mathrm{0},\mathrm{8840238122} \\ $$
Commented by Dwaipayan Shikari last updated on 16/Feb/21
$${Great}\:{sir}! \\ $$
Commented by Dwaipayan Shikari last updated on 16/Feb/21
$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}−{x}}−\frac{\mathrm{1}}{\mathrm{3}+{x}}+… \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{3}+{x}\right)^{\mathrm{2}} }+… \\ $$$$\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}{x}\right){tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }+… \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\pi^{\mathrm{3}} }{\mathrm{8}}.\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }−\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{8}^{\mathrm{3}} }−\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }+… \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{27}.\mathrm{6}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+…\right) \\ $$$$\Rightarrow\frac{\mathrm{4}\pi^{\mathrm{3}} }{\mathrm{81}\sqrt{\mathrm{3}}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+… \\ $$
Commented by mnjuly1970 last updated on 16/Feb/21
$${please}\:{prove}\:{this}\:{nice}\:{identity} \\ $$$${mr}\:{dwaipayan}.. \\ $$$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}\left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}\:+\frac{\mathrm{1}}{\mathrm{3}+{x}}−\frac{\mathrm{1}}{\mathrm{3}−{x}}… \\ $$$${of}\:{course}\:{i}\:{think} \\ $$$$\:{it}\:{can}\:{be}\:{deduced}\:{by} \\ $$$$\:{weiresteress}\:{gamma}\:{formula}.. \\ $$$$\:\Gamma\left({x}\right)=\frac{{e}^{−\gamma{x}} }{{x}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{{e}^{\frac{{x}}{{k}}} }{\mathrm{1}+\frac{{x}}{{k}}} \\ $$$${and}\:{its}\:{corallary} \\ $$$$\:\frac{{sin}\left(\pi{x}\right)}{\pi{x}}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{k}^{\mathrm{2}} }\right)\:.. \\ $$
Commented by Dwaipayan Shikari last updated on 16/Feb/21
$${cosx}=\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}\right)\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{5}\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{5}\pi}\right)… \\ $$$${log}\left({cosx}\right)={log}\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\pi}\right)+{log}\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}\right)+… \\ $$$${Differentiating}\:{both}\:{sides}\:{respect}\:{to}\:{x} \\ $$$$−{tanx}=\frac{\frac{−\mathrm{2}}{\pi}}{\mathrm{1}−\frac{\mathrm{2}{x}}{\pi}}+\frac{\frac{\mathrm{2}}{\pi}}{\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}}−\frac{\frac{\mathrm{2}}{\mathrm{3}\pi}}{\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}\pi}}+.. \\ $$$${tanx}=\frac{\mathrm{1}}{\frac{\pi}{\mathrm{2}}−{x}}−\frac{\mathrm{1}}{\frac{\pi}{\mathrm{2}}+{x}}+\frac{\mathrm{1}}{\frac{\mathrm{3}\pi}{\mathrm{2}}−{x}}−\frac{\mathrm{1}}{\frac{\mathrm{3}\pi}{\mathrm{2}}+{x}}+… \\ $$$$\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}−{x}}−\frac{\mathrm{1}}{\mathrm{3}+{x}}+\frac{\mathrm{1}}{\mathrm{5}−{x}}−\frac{\mathrm{1}}{\mathrm{5}+{x}}+… \\ $$
Commented by mnjuly1970 last updated on 16/Feb/21
$${grateful}\:{mr}\:{Dwaipayan} \\ $$$${thanks}\:{lot}… \\ $$
Answered by mnjuly1970 last updated on 16/Feb/21
$${solution}: \\ $$$$\:\:{S}\:{is}\:{convergent}…\: \\ $$$$\:\:\:{rearrangement}… \\ $$$$\:\:\:{S}=\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }+….\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{3}} }+…\right) \\ $$$$\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{27}}\left\{\underset{{n}=\mathrm{0}\:} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} }−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\:\right)^{\mathrm{3}} }\right\} \\ $$$${note}\:::\:\left(\mathrm{1}\right)::\:\:\:\boldsymbol{\psi}\left({z}\right)−\boldsymbol{\psi}\left(\mathrm{1}−{z}\right)=−\boldsymbol{\pi}{cot}\left(\pi{z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right):\::\:\:\boldsymbol{\psi}'\left({z}\right)+\boldsymbol{\psi}'\left(\mathrm{1}−{z}\right)=\boldsymbol{\pi}^{\mathrm{2}} \left(\mathrm{1}+{cot}^{\mathrm{2}} \left(\pi{z}\right)\right) \\ $$$$\:\left(\mathrm{3}\right)\:::\:\:\boldsymbol{\psi}''\left({z}\right)−\boldsymbol{\psi}''\left(\mathrm{1}−{z}\right)=−\mathrm{2}\boldsymbol{\pi}^{\mathrm{3}} \left(\mathrm{1}+{cot}^{\mathrm{2}} \left(\boldsymbol{\pi}{z}\right)\right){cot}\left(\boldsymbol{\pi}{z}\right) \\ $$$$\:\:\:\:\boldsymbol{\psi}''\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\boldsymbol{\psi}''\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=−\mathrm{2}\pi^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\:\:\boldsymbol{\psi}''\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\boldsymbol{\psi}''\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{−\mathrm{8}\pi^{\mathrm{3}} }{\mathrm{3}\sqrt{\mathrm{3}}}\:…. \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{27}}\left(\frac{−\boldsymbol{\psi}''\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{2}}+\frac{\boldsymbol{\psi}''\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{54}}\left(\frac{\mathrm{8}\pi^{\mathrm{3}} }{\mathrm{3}\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{4}\boldsymbol{\pi}^{\mathrm{3}} }{\mathrm{81}\sqrt{\mathrm{3}}}\:\:….\checkmark\checkmark\:{m}.{n}.. \\ $$$$\:\: \\ $$
Answered by Ajetunmobi last updated on 16/Feb/21
$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sum}}\:=\boldsymbol{\mathrm{S}} \\ $$$$\boldsymbol{\mathrm{seperating}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{negative}}\:\boldsymbol{\mathrm{term}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{factor}}\:\boldsymbol{\mathrm{out}}\:\boldsymbol{\mathrm{minus}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}} \\ $$$$\boldsymbol{\mathrm{S}}=\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\mathrm{3}} }−\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{3}\boldsymbol{\mathrm{k}}+\mathrm{2}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{27}}\left\{\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{k}}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} }−\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{k}}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }\right\} \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\mathrm{1}}{\mathrm{27}}\left\{\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right\}\:\boldsymbol{\mathrm{Answer}} \\ $$$$\boldsymbol{\mathrm{Ajetunmobi}}\:\boldsymbol{\mathrm{Abdulqoyyum}} \\ $$
Answered by Ajetunmobi last updated on 16/Feb/21
$$\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{should}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{tetragamma}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{trigamma}} \\ $$$$\boldsymbol{\mathrm{sorry}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{mistakes}} \\ $$