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

Question

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.

Answer 1

Hint : For $n>1:$

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

Hence?

Answer 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.

Answer 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.

Answer 4

As an alternative since

$$\frac{n}{1+n^2}\sim \frac1n$$

the given series diverges by limit comparison test with $\sum \frac1n$.

Answer 5

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.)

Answer 6

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 $$\left( \begin{array}{ccc} p & \text{approximation} & \text{exact} \\ 10 & 2.25793 & 2.26161 \\ 20 & 2.92608 & 2.92706 \\ 30 & 3.32321 & 3.32366 \\ 40 & 3.60673 & 3.60698 \\ 50 & 3.82737 & 3.82754 \\ 60 & 4.00803 & 4.00814 \\ 70 & 4.16099 & 4.16107 \\ 80 & 4.29363 & 4.29369 \\ 90 & 4.41071 & 4.41077 \\ 100 & 4.51552 & 4.51556 \end{array} \right)$$