How to determine the convergence of the series$\sum_{n=1}^\infty (n^\frac{1}{n^2+1}-1)$? I am currently working on some analysis exercise and was struggling to prove that the series converges $$\sum_{n=1}^\infty (n^\frac{1}{n^2+1}-1)$$
I have tried several tricks such as converting $n^\frac{1}{n^2+1}$ to $e^{logn\frac{1}{n^2+1}}$ and try to apply ratio test and root test to $\sum_{n=1}^\infty( e^{logn\frac{1}{n^2+1}}-1)$ but couldn't succeed.
Also, I tried to upper bound the terms by $n^\frac{1}{n^2}-1$ and try to use a comparison test, but couldn't make it work either, I calculated the sum with Mathematica and it showed me it does converge to approximately $0.87357$, but I just couldn't find a way to prove it.
 A: Using the fact that $e^{x}-1\sim x$ as $ x \to 0$ we can prove convergence by comparison with the series $\sum \frac {\ln n} {n^{2}+1}$. This series is convergent because $\ln n =2\ln \sqrt n\leq  2 \sqrt n$ and $\sum \frac {\sqrt n} {n^{2}+1} \leq \sum \frac {\sqrt n} {n^{2}}  <\infty$.
A: Let $$a_n=b_n-1\qquad \text{with}\qquad b_n=n^\frac{1}{n^2+1}$$
$$\log(b_n)=\frac{\log(n)}{n^2+1}$$ By Taylor expansion or long division
$$\log(b_n)=\frac{\log (n)}{n^2}-\frac{\log
   (n)}{n^4}+O\left(\frac{1}{n^6}\right)$$
$$b_n=e^{\log(b_n)}=1+\frac{\log (n)}{n^2}+\frac{2 \log ^2(n)-4 \log (n)}{4
   n^4}+O\left(\frac{1}{n^6}\right)$$
$$a_n=\frac{\log (n)}{n^2}+\frac{2 \log ^2(n)-4 \log (n)}{4
   n^4}+O\left(\frac{1}{n^6}\right)\tag 1$$ Appply it twice and continue with Taylor
$$\frac{a_{n+1}}{a_n}=1+\frac{1-2 \log (n)}{n^3}+O\left(\frac{1}{n^4}\right)~~<~1$$ So, convergence.
If you use $(1)$ and ignore the high order terms
$$\sum_{n=1}^\infty \frac{\log (n)}{n^2}=\frac{\pi ^2}{6}  (12 \log (A)-\gamma -\log (2\pi ))\sim 0.9375$$ while, without any approximation, the original infinite sum is $\sim 0.9050$
Edit
If we make the expansion of $\log(b_n)$ to order $2k$, we have explicit results for the summation (quite long formulae) and the numerical values are
$$\left(
\begin{array}{cc}
 k & \text{summation} \\
 1 & 0.937548 \\
 2 & 0.901166 \\
 3 & 0.905448 \\
 4 & 0.904966 \\
\infty & 0.904996
\end{array}
\right)$$
A: Comparison can work: let's try to show $\frac{1}{x^p}\ge x^\frac{1}{x^2+1}-1$ for some $1<p$.
$$\frac{1}{x^p}+1\ge x^\frac{1}{x^2+1}$$
$$\log\left(\frac{1}{x^p}+1\right)\ge\frac{\log(x)}{x^2+1}$$
Then by the taylor series for the logarithm we know $\log(1+x)>x-\frac{x^2}{2}$ for $x\in(0,2)$, so we have
$$\frac{1}{x^p}-\frac{1}{2x^{2p}}\ge\frac{\log(x)}{x^2+1}$$
$$\frac{(2x^p-1)(x^2+1)}{2x^{2p}}\ge\log(x)$$
Next, $\sqrt{x}\ge\log(x)$ so it's enough to find $p$ so that
$$\frac{(2x^p-1)(x^2+1)}{2x^{2p}}\ge\sqrt{x}$$
However, notice that the limit of the lhs as $x\to\infty$ is a simple power of $x$: $x^{2-p}$.  Hence if $2-p>\frac{1}{2}$, there will be some finite $N$ after which $x^{2-p}\ge\sqrt{x}$.  The sum up to $N$ converges because it is a finite sum, and the sum after $N$ converges because this inequality shows the terms after $N$ are bounded by the convergent series $\frac{1}{x^p}$.  Any $1<p<\frac{3}{2}$ will work, like say $p=\frac{4}{3}$.
