# Difficult question - convergence of a series

1. Jan 15, 2005

### rachmaninoff

I am trying to show that the series

$$\sum_{n=1}^{\infty}\frac{(\frac{2}{3}+\frac{1}{3}\sin{n})^n}{n}$$ converges (it's a variant of the harmonic series).

So far I got
1) The sequence of partial sums is monotone increasing
2)
$$\frac{1}{3}\leq\frac{2}{3}+\frac{1}{3}\sin{n}\leq1$$
$$\frac{1}{3^n}\leq(\frac{2}{3}+\frac{1}{3}\sin{n})^n\leq1^n$$
$$0<(\frac{2}{3}+\frac{1}{3}\sin{n})^n\leq1$$

Since
$$\sin{x}=1\Rightarrow{x}=\frac{\pi}{2}+2m\pi$$
and $$\pi$$ is irrational
$$\sin{n}<1$$
for all rational or integral n; thus

$$0<(\frac{2}{3}+\frac{1}{3}\sin{n})^n<1\ \forall{n}\in\mathbb{N}$$.

What I've been trying to do here is prove that the sequence of partial sums is Cauchy (edit: which is sufficient to show that it is bounded and the series converges):

$$\sum_{n=j}^{k}\frac{(\frac{2}{3}+\frac{1}{3}\sin{n})^n}{n}<\epsilon$$

But I can't figure out how to show that $$\sin{n}$$ is not arbitrarily close to 1 at some point, and that
$$(\frac{2}{3}+\frac{1}{3}\sin{n})^n$$
is itself not arbitrarily close to 1; thus hindering attempts to make the sum vanish.

I also tried using the power series expansion of $$\sin{n}$$ (which converges everywhere):

$$(\frac{2}{3}+\frac{1}{3}\sin{n})^n=(\frac{2}{3}+\frac{1}{3}(n-\frac{n^3}{3!}+\frac{n^5}{5!}-...))^n=(\frac{2}{3})^n+n(\frac{2}{3})^{n-1}(\frac{n}{3}-\frac{n^3}{3\cdot3!}+...)+...$$
(assuming this is valid, I don't know where it leads)

So what's the trick?

Last edited by a moderator: Jan 15, 2005
2. Jan 15, 2005

### Hurkyl

Staff Emeritus
Okay, the problem you've identified is that you have a subsequence of the integers on which sin n converges to 1.

One thing you might try to do is to find out how frequently sin n is greater than 1 - e (e being small). If it's infrequent enough, you could prove that they don't contribute enough to make the sum diverge.

You might instead consider some special partial sums, like the partial sum of the terms up until the first time sin n > 1-e. Maybe you could prove directly that the partial sums converge.

You might do a transform on your series -- the one I see is grouping the terms in pairs, for instance, then no pair will contain two terms both close to 1, so a bounding argument might work on these.

3. Jan 15, 2005

### Manchot

Well, according to Mathworld, it is unknown whether that particular series converges or diverges. I don't know if you are aware of this or not, and I just wanted to let you know what you're up against.

4. Jan 15, 2005

### rachmaninoff

Aha, an unsolved problem! Figures. :grumpy: