# $\sin n>0$, $a_n=1/n \, ,\sin n<0$, $a_n=-1/n \, , \sum_{n=1}^{\infty} a_n$ converge?

Define sequence $a_n$ as: If $\sin n>0$, $a_n=1/n$, and if $\sin n<0$, $a_n=-1/n$.

Does the series $\displaystyle \sum_{n=1}^{\infty} a_n$ converge?

-
NM............. – Brady Trainor Mar 27 '13 at 8:12
Yes, but I could not apply it here since $\displaystyle \sum_{n=1}^{\infty}|a_n|= \infty$ – math123 Mar 27 '13 at 8:13

Use dirichlets test, which says $$\sum_{i=1}^\infty a_i \epsilon_i$$ converges when $a_i$ is a monotone decreasing sequence converging to zero, and the $$\left|\sum_{i=1}^n \epsilon_i \right|< M$$ for all $n$, (The sum is bounded), where $\epsilon_i$ is $-1$ if $\sin(i)<0$ and $1$ if $\sin(i)>0$.

As $\sin(n)$ is uniform distributed in $(-1,1)$ you should be able to chose $M=10$ (my own guess is that $4$ would be enough too).

-
I don't get how you know M is bounded? You may add 1 many more times than you subtract 1. – Brady Trainor Mar 27 '13 at 8:51
@BradyTrainor Here the choice of signs is according to $\text{sin}$. So you can use the periodicity of the function to show that it is bounded. – awllower Mar 27 '13 at 9:11
In what sense is the sequence $(\sin n)$ uniformly distributed in $(-1, 1)$? Indeed, the fractional part $\{ n/2\pi \}$ of $n / 2\pi$ is equidistributed on $[0, 1]$. But this seemingly strong claim is also insufficient to claim that the partial sum of signs $(\epsilon_i)$ are bounded. So I'm curious if there are some more elaborated results concerning this topic. – Sangchul Lee Mar 27 '13 at 10:23
@sos440 yeah you are right, it was more an heuristic, Does it help to split up the stuff in $$\sign(\sin(n)) \frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{n}} ?$$ It is bounded between -5 and 6 for $n< 10^7$ – Dominic Michaelis Mar 27 '13 at 11:04
@DominicMichaelis, $10^7$ is still infinitesimally small compared to the number of natural numbers, and we do not know how $(s_n)$ will behave for large $n$. Our numerical experiments may help us guess the asymptote, but cannot serve as a proof. – Sangchul Lee Mar 29 '13 at 0:51

Define the sign function $\mathrm{sgn} : \Bbb{R} \to \{-1, 0, 1\}$ as

$$\mathrm{sgn}(x) = \begin{cases} 1 & x > 0 \\ 0 & x = 0 \\ -1 & x < 0 \end{cases}.$$

Then we can rephrase $(a_n)$ as

$$a_n = \frac{\mathrm{sgn}(\sin n)}{n}.$$

Now let $s_n = \sum_{k=1}^{n} \mathrm{sgn}(\sin k)$ be the sum of signs. Then by summation by parts,

\begin{align*} \sum_{k=1}^{n} a_k &= \sum_{k=1}^{n} \frac{s_{k}-s_{k-1}}{k} = \sum_{k=1}^{n} \frac{s_{k}}{k} - \sum_{k=1}^{n-1} \frac{s_{k}}{k+1} \\ &= \frac{s_{n}}{n} + \sum_{k=1}^{n-1} \frac{s_{k}}{k(k+1)}. \end{align*}

From the Weyl's criterion, it follows that

$$\lim_{n\to\infty} \frac{s_n}{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} \mathrm{sgn}(\sin x) \, dx = 0.$$

This shows that $s_n = o(n)$. But this alone is insufficient to prove or disprove that the given series converges or not. Numerical calculations of $(s_n)$ up to $n = 10^7$

shows that the growth of $(s_n)$ is highly suppressed, indicating the possibility that the series converges.

-