$\text{ If }f:R\to R\text{ be a differentiable function at }\mathrm{x}=0\text{ satisfying }f\left(0\right)=0\text{ and }{f}^{\text{'}}\left(0\right)=2,\text{ then }$the value of $\underset{r\to 0}{\lim }\frac{1}{r}\sum _{n=1}^{\infty }{(-1)}^{n}f\left(\frac{r}{n}\right)$is

$\begin{array}{c}\underset{r\to 0}{\lim }\frac{1}{r}\sum _{n=1}^{\infty }{(-1)}^{n}f\left(\frac{r}{n}\right)=\underset{r\to 0}{\lim }\sum _{n=1}^{\infty }\frac{{(-1)}^{n}f\left(\frac{r}{n}\right)}{r}\\ =\underset{r\to 0}{\lim }\sum _{n=1}^{\infty }\frac{{(-1)}^n\frac{1}{n}{f}^{\text{'}}\left(\frac{r}{n}\right)}{1}\\ =\sum _{n=1}^{\infty }\frac{{(-1)}^n\frac{1}{n}{f}^{\text{'}}\left(0\right)}{1}\\ =\underset{r\to 0}{\lim }\sum _{n=1}^{\infty }\frac{{(-1)}^n\left(2\right)}{n}\\ =-2\underset{r\to 0}{\lim }\sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n}\\ =-2{\log }_{e}2\end{array}$