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Question Number 127080 by Dwaipayan Shikari last updated on 26/Dec/20

Σ_(n=1) ^∞ (1/(e^(−φn) +((e^(2πn) −e^(−2φn)  )/(2e^(−φn) +((e^(2πn) −e^(−2φn) )/(2e^(−φn) +((e^(2πn) −e^(−2φn) )/(2e^(−2φn) ...))))))))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} \:}{\mathrm{2}{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\mathrm{2}\phi{n}} ...}}}} \\ $$

Commented by Dwaipayan Shikari last updated on 26/Dec/20

(√a)=(√a)−(√k)+(√k)=(√k)+((a−k)/( (√a)+(√k)))=(√k)+((a−k)/(2(√k)+((a−k)/( 2(√k)+((a−k)/(2(√k)+...))))))  k=e^(−2φn)   a=e^(2πn)   It is (1/( (√k)+((a−k)/(2(√k)+((a−k)/(2(√k)+((a−k)/(...))))))))=(1/( (√a)))⇒(1/(e^(−φn) +((e^(2πn) −e^(−2φn) )/(2e^(−φn) +...))))=e^(−πn)   SoΣ_(n=1) ^∞ (1/( (√a)))=Σ_(n=1) ^∞ e^(−πn) =(1/(e^π −1))

$$\sqrt{{a}}=\sqrt{{a}}−\sqrt{{k}}+\sqrt{{k}}=\sqrt{{k}}+\frac{{a}−{k}}{\:\sqrt{{a}}+\sqrt{{k}}}=\sqrt{{k}}+\frac{{a}−{k}}{\mathrm{2}\sqrt{{k}}+\frac{{a}−{k}}{\:\mathrm{2}\sqrt{{k}}+\frac{{a}−{k}}{\mathrm{2}\sqrt{{k}}+...}}} \\ $$$${k}={e}^{−\mathrm{2}\phi{n}} \:\:{a}={e}^{\mathrm{2}\pi{n}} \\ $$$${It}\:{is}\:\frac{\mathrm{1}}{\:\sqrt{{k}}+\frac{{a}−{k}}{\mathrm{2}\sqrt{{k}}+\frac{{a}−{k}}{\mathrm{2}\sqrt{{k}}+\frac{{a}−{k}}{...}}}}=\frac{\mathrm{1}}{\:\sqrt{{a}}}\Rightarrow\frac{\mathrm{1}}{{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\phi{n}} +...}}={e}^{−\pi{n}} \\ $$$${So}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{a}}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{−\pi{n}} =\frac{\mathrm{1}}{{e}^{\pi} −\mathrm{1}} \\ $$

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