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Question Number 114099 by mnjuly1970 last updated on 17/Sep/20
.... mathematical analysis.... prove that :: Σ_(n=1) ^∞ ( ((3^n −1)/4^n ))ζ(n+1) =π m.n.july.1970#
$$\:\:\:….\:\:{mathematical}\:\:{analysis}….\:\: \\ $$$$\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\:\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{4}^{{n}} }\right)\zeta\left({n}+\mathrm{1}\right)\:=\pi\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970}# \\ $$$$ \\ $$
Answered by maths mind last updated on 17/Sep/20
=Σ_(n≥1) .Σ_(m≥1) (((3^n −1)/4^n )).(1/m^(n+1) ) =Σ_(m≥1) (1/m){Σ_(n≥1) ((3/(4.m)))^n −Σ_(n≥1) ((1/(4m)))^n } =Σ_(m≥1) (1/m){(3/(4m)).(1/(1−(3/(4m))))−(1/(4m)).(1/(1−(1/(4m))))} =Σ_(m≥1) (1/m){(3/(4m−3))}−Σ_(m≥1) (1/(m(4m−1))) =Σ_(m≥0) (3/((1+m)(4m+1)))−Σ_(m≥0) (1/((m+1)(4m+3))) =Σ_(m≥0) ((1−(1/4))/((m+1)(m+(1/4))))−Σ_(m≥0) ((1−(3/4))/((m+1)(m+(3/4)))) =Ψ(1)−Ψ((1/4))−{Ψ(1)−Ψ((3/4))} =Ψ((3/4))−Ψ((1/4))=Ψ(1−(1/4))−Ψ((1/4))=πcot((π/4))=π
$$=\underset{{n}\geqslant\mathrm{1}} {\sum}.\underset{{m}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{4}^{{n}} }\right).\frac{\mathrm{1}}{{m}^{{n}+\mathrm{1}} } \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{m}}\left\{\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{3}}{\mathrm{4}.{m}}\right)^{{n}} −\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{4}{m}}\right)^{{n}} \right\} \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{m}}\left\{\frac{\mathrm{3}}{\mathrm{4}{m}}.\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}{m}}}−\frac{\mathrm{1}}{\mathrm{4}{m}}.\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{m}}}\right\} \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{m}}\left\{\frac{\mathrm{3}}{\mathrm{4}{m}−\mathrm{3}}\right\}−\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{m}\left(\mathrm{4}{m}−\mathrm{1}\right)} \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}}{\left(\mathrm{1}+{m}\right)\left(\mathrm{4}{m}+\mathrm{1}\right)}−\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({m}+\mathrm{1}\right)\left(\mathrm{4}{m}+\mathrm{3}\right)} \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}{\left({m}+\mathrm{1}\right)\left({m}+\frac{\mathrm{1}}{\mathrm{4}}\right)}−\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}}{\left({m}+\mathrm{1}\right)\left({m}+\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$=\Psi\left(\mathrm{1}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\left\{\Psi\left(\mathrm{1}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right\} \\ $$$$=\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\Psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\pi{cot}\left(\frac{\pi}{\mathrm{4}}\right)=\pi \\ $$
Commented by mnjuly1970 last updated on 18/Sep/20
thank you sir.grateful...
$${thank}\:{you}\:{sir}.{grateful}… \\ $$

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