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?
 A: Hint: Write $\displaystyle  \sin(x)-\frac{x}{1+x}=x^2+O(x^3)$.
A: 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 $$ \sum O(\frac{1}{n^{2\alpha-2}}).$$
A: 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?
A: 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 
$$2(\alpha-1)>1\iff\alpha>\frac32.$$
