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Question-110920

Question Number 110920 by Dwaipayan Shikari last updated on 31/Aug/20 Commented by Dwaipayan Shikari last updated on 31/Aug/20 Ihavefoundthiswhileexperimenting$$\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)…