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Question Number 126732 by 0731619177 last updated on 23/Dec/20

Answered by Dwaipayan Shikari last updated on 24/Dec/20

lim_(x→0) [ψ(x)+(1/x)]=−γ+Σ_(n=1) ^∞ (1/n)−(1/((n−1)+x))+(1/x) Σ_(n=1) ^∞ (1/n)−(1/((n−1)+x))∼−(1/x) (x is very small with respect to Σ^∞ (1/n)) lim_(x→0) 1−(1/x)+(1/2)−(1/(1+x))+(1/3)−(1/(2+x))+....=−(1/x) =−γ−(1/x)+(1/x)=−γ

limx→0[ψ(x)+1x]=−γ+∑∞n=11n−1(n−1)+x+1x∑∞n=11n−1(n−1)+x∼−1x(xisverysmallwithrespectto∑∞1n)lim1x→0−1x+12−11+x+13−12+x+....=−1x=−γ−1x+1x=−γ

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