I received a task to find out whether the following series converges:


On first look it seems simple, but as I keep thinking about it, there's not a single lemma or criterion that I can use to tackle the problem.

D'alembert ? Doesn't work: The following is meaningless IMHO: $\lim\limits_{x \to \infty}{}\frac{\sin(x+1)}{\sin(x)}$

That series isn't monotonic... you can't understand the rules for when will a member of the series be negative or positive.

All I know is that $\sin(x)$ is blocked between (-1) and 1.
Though it's easy to see that $ \sum\limits_{x=1}^\infty\lvert\sin(x) \rvert$ diverges.

May I use Leibniz formula for $\pi$ in order to construct 2 subseries:

  • One that shows that $\sin(x)$ converges to the limit 1
  • Another one that shows that $\sin(x)$ converges to the limit 0

And we know that a series can't converge to 2 different numbers, hence it diverges?

  • $\begingroup$ Sorry but would you guys please add a simple proof for why $\sin(x)$ diverges? I'm not (yet) a mathematician that can understand the proof of "partial sums"... $\endgroup$
    – Dor
    Nov 22 '14 at 20:53
  • $\begingroup$ Dear Dor, convergence or divergence of the series is, really, a claim about the sequence of partial sums of the series. It is going to be essentially impossible to explain why the series does not converge without making use of its partial sums! $\endgroup$ Nov 22 '14 at 21:05
  • $\begingroup$ @MarianoSuárez-Alvarez I know, but someone added a proof of partial sums which includes the expression $e^{i(n-1)/2}$, and I don't understand the meaning of that... How do I use it to prove the divergence of $\sin(x)$? It's not as clear to me as it is clear to you... $\endgroup$
    – Dor
    Nov 22 '14 at 21:52

The sequence $(\sin n)$ doesn't converge to $0$ so the given series is divergent.

  • 1
    $\begingroup$ In which sentence/lemma did you use in order to figure it out? $\endgroup$
    – Dor
    Nov 21 '14 at 20:51
  • $\begingroup$ I doubt there's a name for this; the partial sums aren't a Cauchy sequence if the summands don't converge to $0$. Any convergent sequence is Cauchy. $\endgroup$ Nov 21 '14 at 20:53
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    $\begingroup$ If the series $\sum u_n$ converges then its partial sum $S_n$ is convergent sequence and then $u_n=S_n-S_{n-1}$ converges to $0$. Now use the contrapositive. $\endgroup$
    – user63181
    Nov 21 '14 at 20:54
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    $\begingroup$ It's pretty easy to show that $\sin n$ cannot converge to zero. If $\sin n$ is close enough to zero, then $\sin (n+1)$ is close to $\pm \sin 1$. $\endgroup$ Nov 21 '14 at 20:54
  • 3
    $\begingroup$ "The contrapositive is a mathematician's best friend. Or rather, anyone whose best friend isn't the contrapositive, isn't a mathematician." $\endgroup$
    – user18862
    Nov 22 '14 at 7:52

Hint: The partial sums have an explicit form, because there are the imaginary part of some geometric series.

$$ \sum_{k=1}^n e^{ik} = \frac {1-e^{in}}{1-e^i} = \frac {e^{-in/2}-e^{in/2}}{e^{-i/2}-e^{i/2}} \frac{e^{in/2}}{e^{i/2}} = \frac{\sin \frac n2}{\sin \frac 12} e^{i(n-1)/2} $$ so the $n$th partial sum is $$ \frac{1}{\sin \frac 12} \sin \frac n2\sin{\frac{n-1}2} = \frac{1}{2\sin \frac 12} \left(\cos \frac 12 - \cos \frac{2n-1}4\right) $$

From this, you can explicitly see what values are taken by these partial sums.

  • $\begingroup$ You are thus left with showing that $\cos(2n-1)/4$ does no converge, which is almost equivalent to the original problem :-) $\endgroup$ Nov 22 '14 at 8:09
  • $\begingroup$ well, there is some difference between the convergence of $U_n$ and the convergence of $\sum U_n$ $\endgroup$
    – mookid
    Nov 22 '14 at 9:46
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    $\begingroup$ Apparently the response side-steps the issue about the convergence of the sum. So why does it merit +7 points? $\endgroup$
    – M. Wind
    Nov 22 '14 at 18:32
  • $\begingroup$ @M.Wind well, you can downvote it. It's not like I care about your particular opinion. $\endgroup$
    – mookid
    Nov 22 '14 at 18:33
  • $\begingroup$ Downvoting your asnswer would be petty. $\endgroup$
    – M. Wind
    Nov 22 '14 at 20:17

If $\sin(n) \to 0$, then $|\cos(n)| \to 1$, so $\sum \cos(n)$ diverges, but $\sin(n+1)= \sin(n)\cos(1) + \cos(n)\sin(1)$, so $\sum \sin(1) \cos(n) = \sum \sin(n+1) - \sum \cos(1)\sin(n)$.

  • $\begingroup$ Obviously your comment is mathematically sound. It is less clear what the purpose is. The formula you present can be seen as a first step to rearrange terms in the infinite sum. But then what? Are you suggesting that rearranging terms in this way leads to cancellation of terms in the sum or even convergence? If that is not the case, what it is that you want to say? $\endgroup$
    – M. Wind
    Nov 22 '14 at 18:23
  • $\begingroup$ @M.Wind, this answer very clearly proposes a (very nice!) way to answer the question, which is, as you know, «does the series converge?» Quinn is not proposing any rearrangement of anything. $\endgroup$ Nov 22 '14 at 18:57
  • 1
    $\begingroup$ Sorry, but it is not clear to me! And presumably also not helpful for the OP. I don't like it when experts post one formula without comment, thereby implicitly hinting at something. It suggests to me they don't have the guts to state explicitly what they mean. $\endgroup$
    – M. Wind
    Nov 22 '14 at 19:58
  • $\begingroup$ @M.Wind You must meditate on this answer until it becomes clear. Then you shall be enlightened. $\endgroup$ Nov 22 '14 at 23:51

Strictly speaking, the sum does not exist. That is because the sin(n) terms do not converge to zero, but keep oscillating in the interval (-1, +1). Hence all the partial sums also keep oscillating.

On the other hand, it is a close affair. The amplitude of the terms do not converge to zero, but they also do not diverge. More importantly, the fact that we are dealing with an oscillating sequence means that a lot of cancellation takes place. If one can combine neighbouring terms in a clever way, the cancellation might be such that the sum is convergent!

In my opinion this is an example of a (non-convergent) sum that can be made convergent by relative simple mathematical tricks. Instead of combining terms (which might work, I haven't verified it), there is an even simpler trick. Simply multiply each term with a convergence factor exp(-epsilon*n). Since epsilon is very small, the exponential factor is very close to 1 but just a little bit smaller; this is sufficient to make the summation convergent. If we proceed this way, we obtain:

S = (1/2i) * Sum(n=1 to inf) {exp(n*(i-eps)) - exp(n*(-i-eps))}

It is straightforward to perform the summations over n, which are now convergent.

S = (1/2i) * {exp(i-eps)/(1 - exp(i-eps)) - exp(-i-eps)/(1 - exp(-i-eps))}

We are already in the comfortable position that we can take the limit epsilon -> 0. Rearranging terms and using exp(ix) = cos(x) + i*sin(x), we get

S = (1/2i) * {1/(1 - cos(1) - i * sin(1)) - 1/(1 - cos(1) + i * sin(1))}

S = (1/2) * sin(1) / (1 - cos(1))

S = (1/2) * cos(1/2) / sin (1/2)

The calculation has resulted in a clear, unambiguous result for the sum S !

  • 3
    $\begingroup$ You are answering a question different from the one posed by the OP, which was: doe the series converge? $\endgroup$ Nov 22 '14 at 8:07
  • $\begingroup$ My answer to the OP is very clear. Strictly speaking the series is not converging. On the other hand the sum can be determined ! $\endgroup$
    – M. Wind
    Nov 22 '14 at 8:48
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    $\begingroup$ That does not make any sense: it the series does not converge, then it simply does not have a sum. The question was: does the series converge? (And mentioning all sort of perfectly correct theories about divergent series ---which is something you did not do to justfy anything og what you wrote, by the way--- is not going to help, in a context where the question clearly does not have that in mind) $\endgroup$ Nov 22 '14 at 18:55
  • $\begingroup$ The sum over n! diverges. The sum over sin(n) is nearly convergent. A very different story, as any physicist will confirm. Yet you and your math buddies insist on talking in absolutes. Perhaps that is why some of the replies at this site are a bit lacking. $\endgroup$
    – M. Wind
    Nov 22 '14 at 20:26
  • $\begingroup$ @M.Wind You simply gave a completely wrong answer. If a series is divergent there is no sum. The same series may be summable in a more general context (Abel- / Cesaro-Summability for instance) but that is not the same as convergent and the resulting limits are called Abel-sum or Cesaro-sum, not sum. You are writing condescending comments when you are the one who uses dubious (because unjustified) tricks to get some answer to some question, but not the correct answer to the asked question. You should read some introductory mathematical texts wich teach you how to properly do maths. $\endgroup$
    – AlexR
    Nov 26 '14 at 17:21

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