Convergence or not of infinite series: $\sum^{\infty}_{n=1}\frac{n}{1+n^2}$ [duplicate]

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.

Solution 1:

Hint : For $n>1:$

$\dfrac{n}{1+n^2} \gt \dfrac{n}{n^2+n^2} = (1/2)\dfrac{1}{n}.$

Hence?

Solution 2:

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.

Solution 3:

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.