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 .

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    $\begingroup$ Hmm. I think you need a little more reasoning but your argument is basically an argument about absolute summability in disguise. You just didn't explain it quite well enough I suppose. I would have taken a couple of marks off but you'd get most credit. $\endgroup$ Mar 23, 2015 at 2:45
  • $\begingroup$ I know i should have used the Absolute Convergence / Comparison test, but still... can't I say this. (and of course I skimmed the answer here, but not in the exam...) My girlfriend is doing her Doctorate in math, and she think the whole layout is good except for the theorem called -> I should have called the comparison theorem, not the sandwich... but still, i'm looking for a counter example of what i am saying because this question cost me 10% of my exam. ( and I do want to understand, it's not much about the points at this time) $\endgroup$ Mar 23, 2015 at 2:51
  • $\begingroup$ I think what ended up happening was that you sort of combined absolute convergence with the comparison test in your argument. The two are a little related since you often use a comparison to prove absolute convergence. The comparison test would work but not as you've presented it. $\endgroup$ Mar 23, 2015 at 2:54
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    $\begingroup$ Maybe the technical objection is that the squeezing theorem requires the limits on both ends coincide, but in this case you don't have that. $\endgroup$
    – Fan Zheng
    Mar 23, 2015 at 3:05
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    $\begingroup$ Your mistake was in thinking that bounding by two limits is enough. The two limits must be equal for the squeezing to work, otherwise there is some "room left" between them. $\endgroup$
    – user65203
    Mar 23, 2015 at 9:05

4 Answers 4


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.

  • $\begingroup$ Simple, but highly sufficient. Thank you. $\endgroup$ Mar 23, 2015 at 3:14
  • $\begingroup$ I must say I do love your answer and that it stand here, assuming you are looking at the limit, but I totally forgot to write that $ \frac{-1}{n^2} \leg \frac{sin(\frac{-5}{n^2})}{n^2} \leg \frac{1}{n^2} \qquad \forall n \geq \1 $.... So I will take your answer in this condition as not complete. (thks for the reasonning though) pls read the question again. $\endgroup$ Mar 24, 2015 at 1:50
  • $\begingroup$ @SebastienComtois Added a bit more information as per your recent edit. With having noted that condition, it is indeed sufficient to prove the convergence of the series, however there are still missing gaps in the logic which aught be written (despite being trivial). $\endgroup$
    – JMoravitz
    Mar 24, 2015 at 2:17
  • $\begingroup$ so if I understand, there is no way in the known universe, a way to find a series that is "squeezed" the way I said, but still does not converge. ? $\endgroup$ Mar 24, 2015 at 2:23
  • $\begingroup$ @SebastienComtois Not with the bounds on the summands approaching zero as quickly as they did, there are no counterexamples. Again, your note about the summands being bounded on either side was equivalent to showing that the absolute value was bounded by a rapidly decreasing function. By noting that and applying the comparison test, you show that it is absolutely convergent. Had your bounds on the summand not both converged to zero however, there would still have been room for some oscillation. $\endgroup$
    – JMoravitz
    Mar 24, 2015 at 2:27

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.


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.


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.

  • $\begingroup$ You would have to assume further knowledge about the convergence of the original sequence which is originally asked to prove. Thus highly necessary. Accepting truth of the convergence based on axioms you provide would then prove contradictions. $\endgroup$ Mar 23, 2015 at 3:44
  • $\begingroup$ As explicitly shown by JMoravitz $\endgroup$ Mar 23, 2015 at 3:47

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