Convergence or not of infinite series: $\sum^{\infty}_{n=1}\frac{n}{1+n^2}$ How can we prove that the series $\displaystyle \sum^{\infty}_{n=1}\frac{n}{1+n^2}$ is convergent or divergent?
Solution I try:
$$\lim_{m\rightarrow \infty}\sum^{m}_{n=1}\frac{n}{1+n^2}<\lim_{m\rightarrow \infty}\sum^{m}_{n=1}\frac{n}{n^2}$$
Did not know how I can solve that problem from that point.
 A: As this is a series with positive terms, you can use equivalence: $1+n^2\sim_\infty n^2$, so
$$\frac n{1+n^2}\sim_\infty\frac n{n^2}=\frac 1n,$$
and the harmonic series diverges, so the given series diverges.
A: As an alternative since
$$\frac{n}{1+n^2}\sim \frac1n$$
the given series diverges by limit comparison test with $\sum \frac1n$.
A: Another method is the integral test: $$\int_1^\infty\frac{x}{1+x^2}dx=\frac{1}{2}[\ln (1+x^2)]_1^\infty =\infty.$$(Note that $\frac{x}{1+x^2}=\frac{1}{x+1/x}$ is maximised at $x=1$, by the AM-GM inequality.)
A: Hint : For $n>1:$
$\dfrac{n}{1+n^2} \gt \dfrac{n}{n^2+n^2} = (1/2)\dfrac{1}{n}.$
Hence?
A: Just in case you enjoy generalized harmonic numbers.
We could approximate the partiel sums
$$S_p=\sum^{p}_{n=1}\frac{n}{1+n^2}=\sum^{p}_{n=1}\frac{n}{(n+i)(n-i)}=\frac 12 \left(\sum^{p}_{n=1}\frac{1}{n+i} +\sum^{p}_{n=1}\frac{1}{n-i}\right)$$
$$S_p=\frac{1}{2} \left(H_{p-i}+H_{p+i}-H_i-H_{-i}\right)$$ Now, using the asymptotics, we end with
$$S_p=\left(\gamma-\frac{H_i+H_{-i}}{2}
   \right)+\log \left({p}\right)+\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$ The constant terms evaluates as $\approx -0.09465$.
As shown below, this is not too bad
$$\left(
\begin{array}{ccc}
 p & \text{approximation} & \text{exact} \\
 10 & 2.25793 & 2.26161 \\
 20 & 2.92608 & 2.92706 \\
 30 & 3.32321 & 3.32366 \\
 40 & 3.60673 & 3.60698 \\
 50 & 3.82737 & 3.82754 \\
 60 & 4.00803 & 4.00814 \\
 70 & 4.16099 & 4.16107 \\
 80 & 4.29363 & 4.29369 \\
 90 & 4.41071 & 4.41077 \\
 100 & 4.51552 & 4.51556
\end{array}
\right)$$
A: Showing that the series is smaller than a divergent series tells you nothing. You either want to produce a convergent series that it is smaller than, or a divergent series that it is larger than. We can write
$$ \frac{n}{n^2+1} = \frac{1}{n+(1/n)}. $$
Since $n \geq 1$, $\frac{1}{n} \leq 1$, so $n+ \frac{1}{n} \leq n+1$, so
$$ \frac{n}{n^2+1} \geq \frac{1}{n+1}, $$
and the latter is essentially the harmonic series, that diverges.
