Question Number 110920 by Dwaipayan Shikari last updated on 31/Aug/20 Commented by Dwaipayan Shikari last updated on 31/Aug/20 $${I}\:{have}\:{found}\:{this}\:{while}\:{experimenting} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\mathrm{2}^{{n}} }=\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}.\mathrm{2}^{\mathrm{2}} }+….=−{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)={log}\left(\mathrm{2}\right)…
Question Number 110919 by Dwaipayan Shikari last updated on 31/Aug/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{r}} {r}!}\left(\underset{{k}=\mathrm{1}} {\overset{{r}} {\prod}}\left(\mathrm{2}{k}−\mathrm{1}\right)\right)\right) \\ $$ Answered by mnjuly1970 last updated on…
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Question Number 176424 by Mastermind last updated on 18/Sep/22 $$\mathrm{Using}\:\mathrm{perseval}'\mathrm{s}\:\mathrm{Identity} \\ $$$$\mathrm{Evaluate}\::\:\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{x}}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$ Terms of Service Privacy…
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Question Number 45192 by ajfour last updated on 10/Oct/18 $${To}\:{the}\:{developer},\:{of}\:{tinkutara}; \\ $$$${Sir},\:{since}\:{yesterday}\:{the} \\ $$$${app}\:{crashes}\:{when}\:{i}\:{go}\:{for} \\ $$$${any}\:{page}\:{other}\:{than}\:{the} \\ $$$${main}\:'{sort}\:{by}\:{question}\:{id}\:{page}', \\ $$$${please}\:{help},\:{my}\:{phone}\:{is}\:\mathrm{2}{G} \\ $$$${compatible}\:{android}\:{old}\:{version}, \\ $$$${but}\:{the}\:{app}\:{had}\:{been}\:{running}\:{all} \\…
Question Number 45160 by prashant@1234 last updated on 09/Oct/18 $$\mathrm{L}\left\{\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{t}}+\int_{\mathrm{0}} ^{\mathrm{t}} \mathrm{cosht}\:\mathrm{cost}\:\mathrm{dt}\right\} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com