Math 01May 14, 2023 · One min readPitikorn KhlaisamniangStudentShow that the sum of the infinite serieslog2e−log4e+log16e−…+(−1)nlog22ne+…\log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldotslog2e−log4e+log16e−…+(−1)nlog22ne+…is1ln(22)\frac{1}{\ln(2\sqrt{2})}ln(22)1[ logab=c\log_{a}b=clogab=c is equivalent to ac=ba^{c}=bac=b ]