Interesting Limit (27)

first by changing variables from $x\to \dfrac{1}{x}$ : $\lim_{x\to0^+}\frac1{\ln x}\sum_{n=1}^\infty\frac1{n^2x+n} =- \lim_{x\to\infty}\frac1{\ln x}\sum_{n=1}^\infty\dfrac{x}{n^2+nx} =-\lim_{x\to\infty}\frac1{\ln x}\sum_{n=1}^\infty\left(\dfrac{1}{n}-\dfrac{1}{n+x}\right)= -\lim_{x\to\infty}\frac{\gamma+\psi(x+1)}{\ln x}$ notice that the sum is just the defination of the digamma function . we know asymtopically $\psi(x+1) = -\gamma+ H_x\approx \ln x$ plugging this in we have $-\lim_{x\to\infty}\frac{\psi(x+1)}{\ln x} = -\lim_{x\to\infty}\frac{\gamma+\ln x}{\ln x} = \boxed{-1}$

$\sum _{n=1}^{\infty } \frac{1}{n^2 x+n} \Rightarrow \psi ^{(0)}\left(1+\frac{1}{x}\right)+\gamma$

$\underset{x\to 0}{\text{lim}}\frac{\psi ^{(0)}\left(1+\frac{1}{x}\right)}{\log (x)} \Rightarrow -1$