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I have following series $$\sum_{n=1}^\infty \frac{\sin(3n)}{n^4} $$

Then I use comparison test, compare them to $$\sum_{n=1}^\infty \frac{1}{n^4} $$

And conclude they converge.

However, I have this task marked as mistake "You can do comparison test only on positive number series". So are there alternative ways to prove convergence? I have tried symbolab but it also uses comparison test

tortue
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Arnie
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    @imranfat: thats in itself doesn't prove convergence. – Martin Argerami Nov 24 '19 at 00:18
  • @Martin would you mind explaining in wat sense my comment is different from Jose's answer? – imranfat Nov 24 '19 at 02:18
  • @imranfat: your comment seems to imply that you can do comparison directly. I don't think you can avoid going through absolute convergence. For instance if you take $a_n=(-1)^n\frac{\pi^4}{90}$, then $-\sum_n\tfrac1{n^4}\leq \sum_{n=1}^ma_n\leq\sum_n\tfrac1{n^4}$ for all $m$, and still the series does not converge. – Martin Argerami Nov 24 '19 at 02:52
  • Interesting observation. Thanks. – imranfat Nov 25 '19 at 14:11

2 Answers2

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You can compare the series $\displaystyle\sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert$ with $\displaystyle\sum_{n=1}^\infty\frac1{n^4}$ and deduce from this that the series $\displaystyle\sum_{n=1}^\infty\frac{\sin(3 n)}{n^4}$ converges absolutely. Therefore, it converges.

José Carlos Santos
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We can't use direct comparison test directly since the $a_n$ term of the given series oscillates.

What we can use is the absolute convergence criterion that is

$$\sum |a_n|<\infty \implies \sum a_n<\infty$$

and in this case if we consider

$$\sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert$$

we can apply direct comparison test on that since $|a_n| \ge 0$ and we obtain that

$$\sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert \le \sum_{n=1}^\infty \frac{1}{n^4}$$

thus $\sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert$ converges and by absolute convergence criterion also the original series converges.

To summarize the steps for the proof are

$$ \sum_{n=1}^\infty \frac{1}{n^4}<\infty \stackrel{D.C.T.}\implies \sum_{n=1}^\infty\left\lvert\frac{\sin(3 n)}{n^4}\right\rvert <\infty \stackrel{Abs.C.C.}\implies \sum_{n=1}^\infty \frac{\sin(3 n)}{n^4} <\infty$$

user
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