Question Number 34504 by NECx last updated on 07/May/18 | ||
$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+.... \\ $$$${find}\:{the}\:{sum}\:{of}\:{the}\:{series} \\ $$ | ||
Commented by abdo mathsup 649 cc last updated on 07/May/18 | ||
$${the}\:{question}\:\:{is}\:{not}\:{clear}\:{sir}\:{Necx}... \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 07/May/18 | ||
$${log}\left(\mathrm{1}+{t}\right)={t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{4}} }{\mathrm{4}}.... \\ $$$${log}\left(\mathrm{1}−{t}\right)=−{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{4}} }{\mathrm{4}}... \\ $$$${log}\mathrm{2}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}... \\ $$$${put}\:{t}={i} \\ $$$${log}\left(\mathrm{1}+{i}\right)={i}−\frac{{i}^{\mathrm{2}} }{\mathrm{2}}+\frac{{i}^{\mathrm{3}} }{\mathrm{3}}−\frac{{i}^{\mathrm{4}} }{\mathrm{4}}+\frac{{i}^{\mathrm{5}} }{\mathrm{5}}... \\ $$$${pls}\:{recheck}\:{the}\:{probem}\:{pls} \\ $$ | ||