<s>[INST] 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.
 [/INST] 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 $p=\frac{4}{3}$.
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