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The series in question is the following:

$$\sum_{n=1}^{\infty}n^2 \Big(\sin(\frac{1}{n^{\alpha}})-\frac{1}{n^{\alpha}+1}\Big)$$

I want to study for which $\alpha >0$ does the series converge.

I tried the root test to not much avail, after I tried expanding with taylor the sine function (by the way I can taylor expand a function even if the input are only ration numbers of the form $\frac{1}{n^{\alpha}}$ right?) but I still haven't got anywhere.

We have not done the integral test, is that the way to go?

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4 Answers 4

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Hint: Write $\displaystyle \sin(x)-\frac{x}{1+x}=x^2+O(x^3)$.

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Hint. Using, as $x$ tends to $0$, the expansions $$ \sin x = x+o(x^2)$$ $$\frac{1}{1+x}=1-x+O(x^2)$$ you get $$n^2 \Big(\sin(\frac{1}{n^{\alpha}})-\frac{1}{n^{\alpha}+1}\Big)=n^2 \Big(\frac{1}{n^{\alpha}}-\frac{1}{n^{\alpha}}(1-\frac{1}{n^{\alpha}})+O(\frac{1}{n^{2\alpha}})\Big)=O(\frac{1}{n^{2\alpha-2}})$$ Then you are led to consider $$ \sum O(\frac{1}{n^{2\alpha-2}}).$$

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  • $\begingroup$ A small remark: as written, this only will give one side of the answer (a $O(\cdot)$ is only an upperbound): it will enable to say that for $\alpha$ bigger than a threshold the series converges, but not that for $\alpha$ smaller than this threshold it diverges. (For this, you would need a lower bound; e.g., a $\Theta(\cdot)$ or equivalent). $\endgroup$
    – Clement C.
    Nov 27, 2015 at 15:37
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You can indeed do Taylor expansions for $\frac{1}{n^\alpha}$, as long as $\frac{1}{n^\alpha} \xrightarrow[n\to\infty]{} 0$. (That is, for any $\alpha > 0$. Note that the case $\alpha \leq 0$ is straightforward). Doing so, you obtain: $$ \sin \frac{1}{n^\alpha} = \frac{1}{n^\alpha} + o\left(\frac{1}{n^{2\alpha}}\right) $$ and $$ \frac{1}{n^\alpha+1} = \frac{1}{n^\alpha}\frac{1}{1+\frac{1}{n^\alpha}} = \frac{1}{n^\alpha}\left(1-\frac{1}{n^\alpha}+ o\left(\frac{1}{n^{\alpha}}\right)\right) = \frac{1}{n^\alpha}-\frac{1}{n^{2\alpha}}+ o\left(\frac{1}{n^{2\alpha}}\right) $$ so $$ \sin \frac{1}{n^\alpha} - \frac{1}{n^\alpha+1} = \frac{1}{n^{2\alpha}}+ o\left(\frac{1}{n^{2\alpha}}\right) $$ and finally $$ n^2\left(\sin \frac{1}{n^\alpha} - \frac{1}{n^\alpha+1}\right) = \frac{1}{n^{2\alpha-2}} + o\left(\frac{1}{n^{2\alpha-2}}\right) $$

Can you conclude based on this?

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I would use equivalents, via an asymptotic development:

  • $\sin\dfrac1{n^\alpha}=\dfrac1{n^\alpha}+o\Bigl(\dfrac1{n^{2\alpha}}\Bigr)$,
  • $\dfrac1{n^\alpha+1} =\dfrac1{n^\alpha}\biggl(1-\dfrac1{n^\alpha}+\dfrac1{n^{2\alpha}}+o\Bigl(\dfrac1{n^{2\alpha}}\Bigr)\biggr)$.
  • Hence $$\sin\dfrac1{n^\alpha}-\dfrac1{n^\alpha+1} =\dfrac1{n^{2\alpha}}+o\Bigl(\dfrac1{n^{2\alpha}}\Bigr)\sim_\infty\dfrac1{n^{2\alpha}}$$

Thus $$n^2\biggl(\sin\dfrac1{n^\alpha}-\dfrac1{n^\alpha+1}\biggr)\sim_\infty \dfrac1{n^{2(\alpha-1)}}$$ As the given series has positive terms if $n$ is large enough, we conclude it converges if and only if $$2(\alpha-1)>1\iff\alpha>\frac32.$$

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  • $\begingroup$ Why the down vote? $\endgroup$
    – Bernard
    Nov 27, 2015 at 18:49
  • $\begingroup$ This is a clear answer that I understood why the downvote? is it wrong? $\endgroup$
    – Monolite
    Nov 27, 2015 at 19:48
  • $\begingroup$ If I were not sure it's perfectly correct, I wouldn't have posted it. Maybe some don't like using asymptotic analysis (L'Hospital's worshippers? ;o) $\endgroup$
    – Bernard
    Nov 27, 2015 at 19:56
  • $\begingroup$ It is definitely correct, at least as far as I can see. (Incidentally, it is basically the same approach as the two other answers below.) $\endgroup$
    – Clement C.
    Nov 28, 2015 at 13:25

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