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Good evening everyone,

I'd like to discuss with you the following exercise :

$\sum\limits_{n=1}^{\infty} (-1)^{n} \frac{n^{2} +3n - \sin(n)}{n^{4}-\arctan(n^{2})}$

I can prove that $\lim\limits_{x \to \infty} a_{n} = 0$ , where $a_{n} = \frac{n^{2} +3n - \sin(n)}{n^{4}-\arctan(n^{2})}$

But I can't still proove its convergence, I'd have used Leibnitz alternating series test (due to $(-1)^{n}$), but I was unable to say $a_{n+1} \leq a_{n}$.

Maybe I could study the Absolute convergence and then by Comparison test find that converges ?

Any help would be appreciated,

Thanks anyway.

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    $\begingroup$ $-\pi\leq \arctan(x)\leq \pi$ and $-1\leq \sin(x)\leq 1$ for all $x$. Compare how your series acts then to $\sum\dfrac{n^2}{n^4}$ $\endgroup$ – JMoravitz Jun 29 '18 at 20:05
  • $\begingroup$ @JMoravitz this works ? $\endgroup$ – jacopoburelli Jun 29 '18 at 20:07
  • $\begingroup$ With an appropriately phrased argument, yes. $\endgroup$ – JMoravitz Jun 29 '18 at 20:08
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No need for Leibniz: For $n>1,$

$$\left |(-1)^{n} \frac{n^{2} +3n - \sin n}{n^{4}-\arctan(n^{2})}\right| \le \frac{n^{2} +3n +1}{n^{4}-\pi/2}.$$

This looks good, yes? It appears that the terms on the right are like $1/n^2.$ Verify that the terms on the right divided by $1/n^2$ approach $1.$ Since $\sum 1/n^2<\infty,$ the limit comparsion test shows the sum of the terms on the right converges. Hence our series converges absolutely, and therefore converges. (Moral: Don't blindly assume Leibniz is the way to go just because it's got the $(-1)^n$ thing happening.)

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Hint/Walkthrough: Since $\sin(n)$ and $\arctan(n^2)$ are both easily bounded, while $n^2+3n$ and $n^4$ are not, one can prove that your expression $a_n$ is definitely monotonically decreasing after a certain point. Then you can use the Alternating Series test.

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