(1/(πe))+(3/(2π^2 e^2 ))+((11)/(6π^3 e^3 ))+((25)/(12π^4 e^4 ))+((137)/(60π^5 e^5 ))+...=((log(π)+1−log(πe−1))/(πe−1))


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Question Number 134657 by Dwaipayan Shikari last updated on 06/Mar/21

$\frac{1}{π e} + \frac{3}{2 π^{2} e^{2}} + \frac{11}{6 π^{3} e^{3}} + \frac{25}{12 π^{4} e^{4}} + \frac{137}{60 π^{5} e^{5}} + . . . = \frac{l o g (π) + 1 - l o g (π e - 1)}{π e - 1}$
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