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Question Number 131877 by Dwaipayan Shikari last updated on 09/Feb/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n(e^{2\pi n}-1)}$$

Commented by Dwaipayan Shikari last updated on 09/Feb/21

$$Ihavefound$$$$log(\frac{e^{2\pi }}{e^{2\pi }-1}.\frac{e^{4\pi }}{e^{4\pi }-1}.\frac{e^{6\pi }}{e^{6\pi }-1}...)=-log\left({(1,e^{2\pi })}_{\mathrm{\infty }}\right)$$$$\underset{n=0}{\prod ^{\mathrm{\infty }}}(1-{aq}^n)={(a,q)}_{\mathrm{\infty }}$$

Commented by SEKRET last updated on 09/Feb/21

$$Ihavefound$$$$log(\frac{e^{2\pi }}{e^{2\pi }-1}.\frac{e^{4\pi }}{e^{4\pi }-1}.\frac{e^{6\pi }}{e^{6\pi }-1}...)$$$${\mathbf{e}}^{2\mathit{\pi }}>{\mathbf{e}}^{2\mathit{\pi }}-1$$$$+\mathrm{\infty }$$

Commented by Dwaipayan Shikari last updated on 09/Feb/21

$$Butnotsignificantly\frac{e^{2\pi }}{e^{2\pi }-1}=1.0018..$$$$\frac{e^{4\pi }}{e^{4\pi }-1}<1.0018$$$$\frac{e^{6\pi }}{e^{6\pi }-1}<\frac{e^{4\pi }}{e^{4\pi }-1}<1.0018$$$$So\frac{e^{2\pi }}{e^{2\pi }-1}.\frac{e^{4\pi }}{e^{4\pi }-1}.\frac{e^{6\pi }}{e^{6\pi }-1}...=(1.0018)(1.0018-ϵ)(1.00018-2ϵ)...$$$$Soitconvergesϵ=+ve$$

Commented by Dwaipayan Shikari last updated on 09/Feb/21

$$Ifwetakereverse$$$$log(\frac{e^{2\pi }}{e^{2\pi }-1}.\frac{e^{4\pi }}{e^{4\pi }-1}..)=-log((1-\frac{1}{e^{2\pi }})(1-\frac{1}{e^{4\pi }})...)$$$$Whichconverges..$$

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