Series $A= \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$ Consider convergence of series:
$A=\displaystyle \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$
I have a idea
\begin{align*}
a_n&=n^\frac{1}{n^2+1}-1 
\\&=e^{\frac{\ln n}{n^2+1}}-1 \sim b_n= \dfrac{\ln\,n}{n^2+1} \text{ when } n \to \infty
\end{align*}
I wanna consider convergence of series $B=\displaystyle \sum_{n=1}^{\infty}b_n$. I have trouble here.

OP -- Consider convergence of series:
$A=\displaystyle \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$

$$ n^\frac{1}{n^2+1}-1\ =
  \ \frac{n-1}{\sum_{k=0}^{n^2}\,n^{\frac k{n^2+1}}\ }\ < $$
$$ \frac{n-1}{\left(n^2-\lfloor\frac{n^2+1}2\rfloor\right)
     \cdot n^{\frac 12}}\ <
\ \frac{n^{\frac 12}}{n^2-\lfloor\frac{n^2+1}2\rfloor}\ \le $$
$$ \frac 2{n^\frac 32} $$
hence the series is summable (is convergent).
Great
