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.
 A: To help with the convergence of $$B=\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n^2+1}\right)$$ note that $$\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n^2+1}\right)\leq\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n^2}\right)$$.
For the series on the right, we can use the integral test, considering: $$\int_{1}^{\infty}\frac{\ln(x)}{x^2}\;\mathrm{d}x$$
$$=\int_{1}^{\infty}\frac{-\ln\left(\frac{1}{x}\right)}{x^2}\;\mathrm{d}x$$.
Using the substitution $u=\frac{1}{x},\;\frac{\mathrm{d}u}{\mathrm{d}x}=-\frac{1}{x^2}$, we get
$$\int_{1}^{0}\ln(u)\;\mathrm{d}u$$
$$=-\int_{0}^{1}\ln(u)\;\mathrm{d}u$$
which through integration by parts obtains
$$\left[u(1-\ln(u))\right]_0^{1}$$
$$=1-\lim_{u\to 0}\left(u(1-\ln(u))\right)$$
$$=1-\lim_{u\to 0}\left(\frac{1-\ln(u)}{u^{-1}})\right)$$
$$=1-\lim_{u\to 0}\left(\frac{-u^{-1}}{-u^{-2}}\right)\;\;\;\text{(L'Hoptial's Rule)}$$
$$=1-\lim_{u\to 0}\left(u\right)$$
$$=1$$
which is finite. Hence, by the integral test, $$\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n^2}\right)$$ converges and therefore by the comparison test, your series $$B=\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n^2+1}\right)$$ also converges. Hope this helps as a hint.
A: 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
