Show that $\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right)$ 
Show that $\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right)$.

In fact, this result is pretty obvious but when I did this in homework I basically got no points because I was to shallow in my reasoning. So here is my new attempt.

Proof:
We know that the Taylor series at point $0$ of $\sin(x)$ exists so the "tail" $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ of it also exists.
Basically, we now want to show that there exist a constant $C>0$ and an index $n_0$ such that for all $n>n_0$ it holds: $\Big|\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|\leq C \cdot \frac{1}{n^5}$ which is equivalent to saying $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=O\left(\frac{1}{n^5}\right)$.
For this we show that $\lim\limits_{n\to\infty}n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ converges.
Because $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}$ converges we are allowed to perform the multiplication with a fixed $n$: $n^5\cdot \sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$.
Now the question is what happens to $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$ if $n\to\infty$? Again, this seems very obvious as $n$ only appears in the denominators but my tutor said that this is not a rigorous argument. So I guess that we must find a closed form expression for $\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}$ or we must at least find an upper bound which attains a closed form expression. I would simply take a geometric series $\sum\limits_{k=1}^{\infty}\frac{1}{n^k}=\frac{1}{n-1}$ as $\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}\leq\frac{1}{n^k}$ holds for all $n$ and $k$.
Hence, for all $n$ we have:
$$
0\leq n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k-4}}\leq \frac{1}{n-1}\\\implies \lim\limits_{n\to\infty}n^5\cdot\sum\limits_{k=2}^{\infty}\frac{1}{(2k+1)!~~n^{2k+1}}=0.
$$
So for a constant $C>0$ we find an index $n_0$ which is large enough such that for all $n>n_0$:
$$\Big|n^5\cdot\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|<C \\\implies \Big|\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}\Big|<C \frac{1}{n^5}\\\implies 
\sum\limits_{k=2}^{\infty}\frac{(-1)^k}{(2k+1)!~~n^{2k+1}}=O\left(\frac{1}{n^5}\right)\\\implies
\sin\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3!~n^3}+O\left(\frac{1}{n^5}\right).
$$
Is this correct?
 A: Comment on the Approach in the Question
It would be more rigorous to cite Dominated Convergence as a reason that
$$
\lim_{n\to\infty}n^5\sum_{k=2}^\infty(-1)^k\frac1{(2k+1)!}\frac1{n^{2k+1}}=-\frac1{120}
$$
thereby bounding the remainder of the series.
One could also bound by a geometric series:
$$
\begin{align}
\left|\sum_{k=2}^\infty(-1)^k\frac1{(2k+1)!}\frac1{n^{2k+1}}\right|
&\le\sum_{k=2}^\infty\frac1{120n^{2k+1}}\\
&=\frac1{120\!\left(n^5-n^3\right)}
\end{align}
$$
However, both of these approaches require that you have either defined $\sin(x)$ as its Taylor series, or that you've shown that the Taylor series converges, which seems to be doing more than this question asks, so circularity is a concern.

Another Approach
Integrate by parts $4$ times:
$$
\begin{align}
\sin(x)
&=\int_0^x\cos(t)\,\mathrm{d}t\\
&=x-\int_0^x(x-t)\sin(t)\,\mathrm{d}t\\
&=x-\frac12\int_0^x(x-t)^2\cos(t)\,\mathrm{d}t\\
&=x-\frac16x^3+\frac16\int_0^x(x-t)^3\sin(t)\,\mathrm{d}t\\
&=x-\frac16x^3+\frac1{24}\int_0^x(x-t)^4\cos(t)\,\mathrm{d}t\\
&=x-\frac16x^3+O\!\left(x^5\right)
\end{align}
$$
Since
$$
\begin{align}
\left|\int_0^x(x-t)^4\cos(t)\,\mathrm{d}t\right|
&\le\left|\int_0^x(x-t)^4\,\mathrm{d}t\right|\\
&=\frac15|x|^5
\end{align}
$$
