4
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by user99914, Martin Sleziak, user416281, Jose Arnaldo Bebita-Dris, ncmathsadist Jun 18 '18 at 18:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ limit-comparison? $\endgroup$ – Lord Shark the Unknown Jun 8 '18 at 19:23
  • 1
    $\begingroup$ The right side is not converging, it's the harmonic sum. If you want to compare to the right side, you need to have l.h.s>=r.h.s to conclude it's not converging. How can you modify your inequality so it will be true? $\endgroup$ – JonesY Jun 8 '18 at 19:24
  • $\begingroup$ lord Shark explain please $\endgroup$ – jacky Jun 8 '18 at 19:27
  • 1
    $\begingroup$ As yourself: What, in broad terms, does the $n$th term look like for large $n?$ That $1$ downstairs is minute compared to large $n$ So forget about the $1.$ Then you have $n/n^2= 1/n$ and $\sum 1/n=\infty.$ Now that's just for intuition. Go with it towards a proof. $\endgroup$ – zhw. Jun 8 '18 at 19:29
  • 1
    $\begingroup$ @WeatherVane You mean the sequence is converging, or the series? $\endgroup$ – aschepler Jun 9 '18 at 2:39
19
$\begingroup$

Hint : For $n>1:$

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

Hence?

$\endgroup$
10
$\begingroup$

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 $n \geq 1$, $1/n \leq 1$, so $n+1/n \leq n+1$, so $$ \frac{n}{n^2+1} \geq \frac{1}{n+1}, $$ and the latter is essentially the harmonic series, that diverges.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

As an alternative since

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

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

$\endgroup$
4
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ This would be my answer of choice... $\endgroup$ – imranfat Jun 9 '18 at 16:32
2
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.