Question Number 137347 by Dwaipayan Shikari last updated on 01/Apr/21
$$\frac{\mathrm{1}}{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+...}}}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+...}}}=\frac{\mathrm{1}}{\pi}\:\: \\ $$$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$
Answered by MJS_new last updated on 01/Apr/21
![t=1+(((π+1)^2 )/(1+(((π+2)^2 )/(1+...))))∈R [obviously] (1/(1+(π^2 /t)))+(1/t)=(1/π) but this has no real solution for t ⇒ wrong](Q137354.png)
$${t}=\mathrm{1}+\frac{\left(\pi+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\left(\pi+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}+...}}\in\mathbb{R}\:\left[\mathrm{obviously}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{{t}}}+\frac{\mathrm{1}}{{t}}=\frac{\mathrm{1}}{\pi} \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:{t}\:\Rightarrow\:\mathrm{wrong} \\ $$
Commented by Dwaipayan Shikari last updated on 01/Apr/21
$${Thanks}\:{sir}\:! \\ $$