<s>[INST] On an exercise with a series: $\sum_{n=1}^{\infty}n^2 \Big(\sin(\frac{1}{n^{\alpha}})-\frac{1}{n^{\alpha}+1}\Big)$ 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?
 [/INST] 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?
</s>