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](https://www.tinkutara.com/question/Q126767.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\psi\left({x}\right)+\frac{\mathrm{1}}{{x}}\right]=−\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)+{x}}+\frac{\mathrm{1}}{{x}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)+{x}}\sim−\frac{\mathrm{1}}{{x}}\:\:\left({x}\:{is}\:{very}\:{small}\:{with}\:{respect}\:{to}\:\overset{\infty} {\sum}\frac{\mathrm{1}}{{n}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}1}−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}+{x}}+….=−\frac{\mathrm{1}}{{x}} \\ $$$$=−\gamma−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}}=−\gamma \\ $$