I received a task to find out whether the following series converges:
$$\sum_{x=1}^\infty\sin(x)$$
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?
$\endgroup$ 34 Answers
$\begingroup$The sequence $(\sin n)$ doesn't converge to $0$ so the given series is divergent.
$\endgroup$ 8 $\begingroup$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.
$\endgroup$ 5 $\begingroup$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 $\sum \sin(1) \cos(n) = \sum \sin(n+1) - \sum \cos(1)\sin(n)$.
$\endgroup$ 4 $\begingroup$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 !
$\endgroup$ 6
