Counter-example for misused theorem in series convergence test I had a math exam not so long ago and got my result back, I'm happy with the result but there is a question for which my teacher gave me an explanation (for me not having the points) but I still think my reasoning is good. So I am asking for a justification ( / counter-example).
It's concerning a test of convergence for the series:
$$\sum_{n=1}^{\infty} \frac{\sin( \frac{-5}{n^2})}{n^2} $$
Which I said is converging...

I started by saying that:
$$ -1 \leq  \sin(x) \leq  1$$
And that:
$$\frac{-1}{n^2} \leq  \frac{sin(\frac{-5}{n^2})}{n^2} \leq  \frac{1}{n^2} \qquad \forall n \geq 1$$
Then saying that since we know:
$$ \sum_{n=1}^{\infty} \frac{-1}{n^2} \rightarrow \text{Converge} $$
$$ \sum_{n=1}^{\infty} \frac{1}{n^2} \rightarrow \text{Converge} $$
using the "squeeze theorem" (Note that my exam is in French, so i said the "sandwich theorem")
$$ \sum_{n=1}^{\infty} \frac{-1}{n^2} \leq  \sum_{n=1}^{\infty} \frac{\sin( \frac{-5}{n^2})}{n^2} \leq  \sum_{n=1}^{\infty} \frac{1}{n^2}
$$
We can conclude that the series converge.

My teacher is saying that i can't use this theorem here, and I'm saying that since     both are converging, the middle one as no where to go but to converge...what ever the     value it converge to.
Why can't i use this to prove it's converging.
Is there any counter-example to what i am saying ?
Yes, I know the theorem called is not the good one, but the point here is that i am still comparing the functions on every term $( \forall n \geq 1)$. So the counter example must take that in account.
It is Not a comparison of the limit, neither a squeeze theorem in the sense that i don't want to get a result (no value), but simply to imply that the series is converging .
 A: Firstly, it is highly worthy of note that your argument would be correct if you just noted that the absolute value of $\frac{\sin\left(\frac{-5}{n^2}\right)}{n^2}$ is bounded above by $\frac{1}{n^2}$, the sum of which converges absolutely - having absolute values bounded by an absolutely convergent series is sufficient. Essentially, the fact that you included both positive and negative bounds is sufficient to establish the desired fact. Your answer is definitely very close to being correct, in any case and surely captures the spirit of the answer.
In fact, the statement that you invoke:

Let $f(n)\leq g(n)\leq h(n)$. If $\sum f(n)$ and $\sum h(n)$ converge, then so does $\sum g(n)$.

is true, though not trivial. To sketch a proof, letting $F(n)$, $G(n)$, and $H(n)$ be the partial sums of the sequences, we easily show that if $a>b$ then $F(a)-F(b)<G(a)-G(b)<H(a)-H(b)$. Since the outer two series converge, for large enough $a$, we can force the $F(a)-F(b)$ and $H(a)-H(b)$ terms to be arbitrarily close to $0$, implying the same for $G(a)-G(b)$ - which means $G(n)$ is a Cauchy sequence and thus convergent.
The issue with your reasoning is that we are looking at the convergence of partial sums - the squeeze theorem would apply if you could bound your series by two series converging to the same value - however, since they don't converge to the same value, you've merely bounded the partial sums into some interval - we still need to worry about the possibility of the partial sums oscillating. Though this doesn't happen, this fact is not the squeeze theorem or even a consequence of it - and if you really want the general case, it needs proof.
A: What you say is not true. The fact that a sequence $\{x_n\}$ is between two convergent sequences $\{y_n\}$ and $\{z_n\}$ doesn't imply that the sequence $\{x_n\}$ converges unless both, $\{y_n\}$ and $\{z_n\}$ converges to the same limit.
I think that what you really wanted to say was something like this:
For every $n\in \mathbb{N}:$
$$\left|\sum_{k=1}^{n} \frac{sin( \frac{-5}{k^2})}{k^2}\right|\leq \sum_{k=1}^{n} \left|\frac{sin( \frac{-5}{k^2})}{k^2}\right|\leq \sum_{k=1}^{n}\left|\frac{1}{k^2}\right|\leq \sum_{k=1}^{\infty}\left|\frac{1}{k^2}\right|\leq K$$
so $\sum_{k=1}^{n} \left|\frac{sin( \frac{-5}{k^2})}{k^2}\right|$ is bounded from above and therefore $\sum_{k=1}^{\infty} \left|\frac{sin( \frac{-5}{k^2})}{k^2}\right|$ converges, i.e., $\sum_{k=1}^{\infty} \frac{sin( \frac{-5}{k^2})}{k^2}$ converges absolutely and therefore, in particular, it converges.
A: As mentioned in the comments and in the other answers, the issue with your answer is that although it proves that it does not diverge to infinity, it might still diverge.
As for a counterexample, It can be shown that $\sum\limits_{n=1}^\infty \frac{1}{n^2}= \frac{\pi^2}{6}>1$
The summation $\sum\limits_{n=1}^{\infty} (-1)^n$ diverges since the sequence of partial sums alternates between $-1$ and $0$, never stopping on a single repeated entry.
So:  $\sum\limits_{n=1}^\infty\frac{-1}{n^2}\leq\sum\limits_{n=1}^K \frac{-1}{n^2}\leq \sum\limits_{n=1}^K (-1)^n\leq \sum\limits_{n=1}^K\frac{1}{n^2}\leq\sum\limits_{n=1}^\infty \frac{1}{n^2}$ for all $K$, and the series on the left and on the right both converge, however the one in the center does not.

In regards to your edit and recent comment:  you add that you had noted on your exam that $-\frac{1}{n^2}\leq \frac{\sin(\frac{-5}{n^2})}{n^2}\leq \frac{1}{n^2}$ for each $n$.
In that case, then that precisely the same thing as saying that $|\frac{\sin(\frac{-5}{n^2})}{n^2}|\leq \frac{1}{n^2}$ for each $n$, which is what you wanted to do.  In doing so, you can continue by noting that $\sum\limits_{n=1}^K |\frac{\sin(\frac{-5}{n^2})}{n^2}|\leq \sum\limits_{n=1}^K \frac{1}{n^2}$ for each $K$ (since term-by-term the summands are comparable).
Noting that the series with absolute values has strictly positive entries, it follows that $\sum\limits_{n=1}^\infty|\frac{\sin(\frac{-5}{n^2})}{n^2}|\leq \sum\limits_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ and so $\sum\limits_{n=1}^\infty|\frac{\sin(\frac{-5}{n^2})}{n^2}|$ converges.  (This follows from theorem since for any series where all summands are positive, the sequence of partial sums is monotonically increasing.  Any monotonic bounded sequence is convergent)
By theorem, any absolutely convergent series will also be convergent (see a proof here).  Therefore $\sum\limits_{n=1}^\infty\frac{\sin(\frac{-5}{n^2})}{n^2}$ converges as well.
In fact, for any $f(n)$ such that $-\frac{1}{n^2}\leq f(n)\leq \frac{1}{n^2}$ for all $n$ the proof is identical and will also have $\sum\limits_{n=1}^\infty f(n)$ be convergent.
Even so, if I were the one grading, I would be expecting to hear the statements of the theorems used (that positive-term series bounded above by a convergent series is convergent, and that absolutely convergent series are convergent).
You could forgo some of the trouble of this by noting that $\sin(\frac{-5}{n^2})$ is always negative and so $-1\cdot\sum\limits_{n=1}^\infty \frac{\sin(\frac{-5}{n^2})}{n^2} = \sum\limits_{n=1}^\infty -\frac{\sin(\frac{-5}{n^2})}{n^2}\leq \sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ since $-\frac{\sin(\frac{-5}{n^2})}{n^2}$ is always positive, skipping the necessity of using the linked theorem.
A: Using the axioms you used above, you can not conclude that the original sequence converges. Very simpley $\sum_{n=0}^\infty 1/n^2$ is positive while $\sum_{i=0}^\infty -1/n^2 $ is negative. This allows hypothetically that the sequence could 'oscilate' in this region and thus not converge. You have provided nothing to show that this does not happen.
