0
$\begingroup$

$$ \sum_{n=1}^\infty \dfrac{(-1)^n n^3+n^2}{n^4+1}$$

The difficulty for me is that $(-1)^n$ is just on $n^3$ not on the entire term.

I tried to use $\dfrac{a_{n+1}}{a_n}$, and compare it with other series, but they don't work (probably since the terms of this series are not always positive). Thus I think that in order to deal with this series, maybe I have to use its own property. However, I'm not able to find useful property from it. Thus, any help on this? Thanks!

$\endgroup$
6
  • $\begingroup$ en.wikipedia.org/wiki/Alternating_series_test $\endgroup$
    – Joe Shmo
    Mar 27, 2022 at 21:55
  • $\begingroup$ I tried to use the Leibniz criterion, but the thing is that $(-1)^n$ is hard to deal with. If the series is $\sum_{n=1}^\infty\ \ (-1)^n\ \ \dfrac{ n^3+n^2}{n^4+1}$, then I can use the criterion. Thus the problematic thing for me is the $(-1)^n$ is only on $n^3$, not the entire term... $\endgroup$
    – M_k
    Mar 27, 2022 at 21:58
  • $\begingroup$ Actually you might have a typo in the original question. I think the series as you put it in the comments is the intended question, in which case just note that the sequence $\Big ( \dfrac{n^3 + n^2}{n^4 + 1} \Big)_{n \ge 1}$ decreases monotonically to $0$, whence the alternating series test applies. $\endgroup$
    – Joe Shmo
    Mar 27, 2022 at 22:01
  • $\begingroup$ I hope the question will ask me $\sum_{n=1}^\infty\ (-1)^n\ \dfrac{ n^3+n^2}{n^4+1}$, but unfortunately, I'm pretty sure that my original question is right $\endgroup$
    – M_k
    Mar 27, 2022 at 22:02
  • 2
    $\begingroup$ Simply split the terms: the part with $(-1)^n$ converges by Leibniz, while $$\frac{n^2}{n^4+1} < \frac{1}{n^2}$$ you get a sum of two convergent series, hence it is convergent $\endgroup$
    – Crostul
    Mar 27, 2022 at 22:07

3 Answers 3

4
$\begingroup$

$$\frac{(-1)^n n^3+n^2}{n^4+1}=\underbrace{\frac{(-1)^n(n^3+\frac 1 {n})}{n^4+1}}_{=\frac{(-1)^n}{n}}-\underbrace{\frac{(-1)^n}{n}\frac 1 {n^4+1}}_{|.|\leq \frac 1 {n^4}}+\underbrace{\frac{n^2}{n^4+1}}_{\leq \frac 1 {n^2}}$$

Summing the first term will converge because it's an alternating series.

Summing the other terms will converge because they are dominated by converging series.

$\endgroup$
3
$\begingroup$

First observe that $\dfrac{n^2}{n^4 + 1} \le \dfrac{1}{n^2}$ whence $$ \sum_{n=1}^\infty \dfrac{n^2}{n^4 + 1} \le \sum_{n=1}^\infty \dfrac{1}{n^2} < \infty $$

Likewise, since $\dfrac{n^3}{n^4+1} \searrow 0$, we have

$$ \sum_{n=1}^\infty (-1)^n \dfrac{n^3}{n^4+1} < \infty $$

by the alternating series test. Now,

$$ \sum_{n=1}^\infty \dfrac{(-1)^n n^3+n^2}{n^4+1} = \sum_{n=1}^\infty (-1)^n \dfrac{n^3}{n^4+1} + \sum_{n=1}^\infty \dfrac{n^2}{n^4 + 1} < \infty $$

where the series on the left may be written as the sum of the two series on the right since the latter two converge.

$\endgroup$
3
  • $\begingroup$ @ClementC. I'm not really sure what that is. That wasn't the argument. The argument was that $a_n \sim b_n \implies \sum a_n < \infty \iff \sum b_n < \infty$, but that doesn't apply to alternating series, as somebody in the comments pointed out. $\endgroup$
    – Joe Shmo
    Mar 28, 2022 at 14:32
  • $\begingroup$ we're arguing a moot point, but then the two relevant series would be $\sum \frac{1}{n} + \frac{(-1)^n}{\sqrt{n}}$, and $\sum \frac{1}{n}$, both of which diverge, which makes it an example, not a counterexample. $\endgroup$
    – Joe Shmo
    Mar 28, 2022 at 22:04
  • $\begingroup$ I had ade a mistake in my counterxample (the $\sqrt{n}$ and $n$ should have been swapped for it to be a valid counterexample. Anyway, you fixed your answer now... $\endgroup$
    – Clement C.
    Mar 28, 2022 at 22:46
2
$\begingroup$

Start with noticing that

\begin{align*} \sum_{n=1}^{\infty}\frac{(-1)^{n}n^{3} + n^{2}}{n^{4} + 1} & = \sum_{n=1}^{\infty}\frac{(-1)^{n}n^{3}}{n^{4} + 1} + \sum_{n=1}^{\infty}\frac{n^{2}}{n^{4} + 1} \end{align*}

The second series converges due to the comparison test (why?).

The first series also converges due to the Leibniz test, as the general term is decreasing: \begin{align*} \frac{(n + 1)^{3}}{(n + 1)^{4} + 1} - \frac{n^{3}}{n^{4} + 1} & = \frac{(n + 1)^{3}(n^{4} + 1) - (n + 1)^{4}n^{3} - n^{3}}{((n + 1)^{4} + 1)(n^{4} + 1)}\\\\ & = \frac{(n + 1)^{3}(n^{4} - (n + 1)n^{3}) + (n + 1)^{3} - n^{3}}{((n + 1)^{4} + 1)(n^{4} + 1)}\\\\ & = \frac{-(n + 1)^{3}n^{3} + (n + 1)^{3} - n^{3}}{((n + 1)^{4} + 1)(n^{4} + 1)}\\\\ & = \frac{(n + 1)^{3}(1 - n^{3}) - n^{3}}{((n + 1)^{4} + 1)(n^{4} + 1)} < 0 \end{align*}

Hopefully this helps!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .