How do I determine the convergence of $\sum_{n=1}^\infty (-1)^n n\sin(\frac{1}{n^2})$ I tried to prove whether $\sum_{n=1}^\infty (-1)^n n\sin(\frac{1}{n^2})$ converges or not for some long time.
Of course it is alternating, but the positive term is not monotonically decreasing. Liouville's theorem implies that $n\sin(\frac{1}{n^2})\leq \frac{A}{n}$ for some positive constant $A$. But this is not relevant to determine the convergence of the series. Even though it has a bound, it need not be monotonic.
How do I prove this? Thank you in advance.
 A: Let us look at the behaviour of $n\sin(1/n^2)$ for large $n$. 
Let $f(t)=\frac{\sin(t^2)}{t}$.
We have 
$$f'(t)=\frac{2t^2\cos(t^2)-\sin(t^2)}{t^2}.$$
The numerator can be rewritten as $\frac{1}{t^2\cos^2 t}\left(2-\frac{\tan(t^2)}{t^2}\right)$. 
Since $\frac{\tan u}{u}\to 1$ as $u\to 0$, it follows that for small positive $t$ the function $f'(t)$ is positive. 
So for large enough $n$ the function $n\sin(1/n^2)$ is decreasing, and we can use the alternating series test.
A: You have 
$$
\sin\frac1{n^2}=\frac1{n^2}+\frac{\cos(c(n))}6\,\frac1{n^6},
$$
where $c(n)$ is a number between $0$ and $1/n^2$.
Then
$$
\sum_{n=1}^\infty (-1)^n n\sin(\frac{1}{n^2})
=\sum_{n=1}^\infty (-1)^n n\,\frac1{n^2}+\sum_{n=1}^\infty (-1)^n n \,\frac{\cos(c(n))}6\,\frac1{n^6}\\
=\sum_{n=1}^\infty (-1)^n \,\frac1{n}+\sum_{n=1}^\infty (-1)^n \,\frac{\cos(c(n))}{6n^5}.
$$
The two series on the right converge, so the series on the left converges. 
A: $x - \frac{1}{6} x^3 \le \sin(x) \le x$ for any $x \ge 0$  [which can be proven by differentiation].
$\def\nn{\mathbb{N}}$
Thus $n \sin(\frac{1}{n^2}) \in \frac{1}{n} + [\frac{1}{6n^5}]$  [where "$[r]$" denotes "$\{ x : x \in \mathbb{R} \land |x| \le |r| \}$"].
Thus $\sum_{n=1}^m (-1)^n n \sin(\frac{1}{n^2}) \in \sum_{n=1}^m \left( (-1) \frac{1}{n} + [\frac{1}{6n^5}] \right)$
$\quad \subseteq \sum_{n=1}^m (-1) \frac{1}{n} + [ \sum_{n=1}^m \frac{1}{6n^5} ]$ which is bounded over all $m \in \nn^+$
Thus $\left( \sum_{n=1}^m (-1) n \sin(\frac{1}{n^2}) \right)_{m \in \nn}$ has a convergent subsequence
Thus $\sum_{n=1}^\infty (-1) n \sin(\frac{1}{n^2})$ converges since each term goes to $0$ as $n \to \infty$.
A: For $0<x\leq 1$ we have $$x> \sin x =[x- \frac {x^3}{3!}]+ [\frac {x^5}{5!}-\frac {x^7}{7!}] +[\frac {x^9}{9!}-\frac {x^{11}}{11!}]+... >x-\frac {x^3}{3!}.$$ So for $n\in N$ we have $$n\sin [n^{-2}]> \frac {1}{n}-\frac {1}{3!n^5}>\frac {1}{n+1}>(n+1)\sin [(n+1)^{-2}]>0.$$ So we have an alternating series of terms whose absolute values are monotonically decreasing to $0$.
We can obtain $\;[x>0\implies \sin x>x-x^3/3!]\;$ without using the whole power series, as follows:$$(1)...x>0\implies x>\sin x\implies x^2/2=\int_0^x t\;dt>\int_0^x \sin t\;dt=1-\cos x.$$ $$ \text {(2)... Therefore } x>0 \implies   x^3/3!=\int_0^x (t^2/2)\;dt>\int_0^x(1-\cos t)\;dt=x-\sin x.$$
