Menu Close

Question-126732




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
limx0[ψ(x)+1x]=γ+n=11n1(n1)+x+1xn=11n1(n1)+x1x(xisverysmallwithrespectto1n)lim1x01x+1211+x+1312+x+.=1x=γ1x+1x=γ

Leave a Reply

Your email address will not be published. Required fields are marked *