Testing the series $\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}$

Solution 1:

Ross' handwaving can be justified.

We use Weyl's theorem on the equidistribution of polynomials (mod 1) with at least one irrational coefficient (see here: How do you prove that $p(n \xi)$ for $\xi$ irrational and $p$ a polynomial is uniformly distributed modulo 1?)

Choose $\displaystyle p(x) = 2\pi x + \pi$.

Let $\displaystyle k \in (0,2)$. Let $\displaystyle \delta = k$ if $\displaystyle k \in (0,1)$ and $\displaystyle \delta = k-1$ otherwise.

By Weyl's theorem, density of $n$ such that the fractional part of $\displaystyle p(n) < 1- \delta$ is non zero.

Thus the density (among the set of natural numbers) of the odd numbers $\displaystyle 2n+1$ for which the fractional part of $\displaystyle (2n+1)\pi < 1 - \delta$ is non-zero. Thus the density of $\displaystyle [(2n+1)\pi]$ is non-zero too.

For such an $\displaystyle n$,

Let $\displaystyle N = [(2n+1)\pi]$.

We have that $\displaystyle (2n+1)\pi - N < 1 - \delta$ and so $\displaystyle \cos(N) - \cos((2n+1)\pi) < 1 - \delta$

And thus $\displaystyle \cos(N) < -\delta$ and so $\displaystyle \cos(N) + k < k - \delta \leq 1$

Thus for we have a positive density subset $\displaystyle S$ of $\displaystyle \mathbb{N}$ for which $\displaystyle \cos(N) + k < 1$.

We have that

$$ \sum_{N \in S} \frac{1}{N^{k+\cos(N)}} > \sum_{N \in S} \frac{1}{N}$$

Since $\displaystyle S$ is of positive density, the sum of reciprocals is divergent, for a proof see here: Theorem on natural density

I will leave my earlier attempt here.

For $k \in (0,1)$ consider the below.

Lemma:

Given any $\displaystyle k \in (0,1)$, we can find an infinite number of odd numbers $\displaystyle M$ such that $\displaystyle M\pi - [M\pi] < 1 - k$.

Proof:

This easily follows from Vinogradov's theorem that $\displaystyle \\{p_{n}\alpha\\}$ is equidistributed for irrational $\displaystyle \alpha$ where $\displaystyle p_{n}$ is the $\displaystyle n^{th}$ prime. (See here: http://en.wikipedia.org/wiki/Equidistributed_sequence)

$\displaystyle \circ$

For $\displaystyle n = [M\pi]$ we have

$\displaystyle \cos(n) - \cos(M\pi) = \cos(n) + 1 < 1- k$

(using $\displaystyle |\cos x - \cos y| < |x-y|$).

And so $\displaystyle \cos(n) + k < 0$ for an infinite number of $\displaystyle n$ and so the series $\displaystyle \sum \frac{1}{n^{k+\cos n}}$ must be divergent.

For the $\displaystyle \sin n$ case, we need the sequence $\displaystyle \\{(4k+3)\pi/2\\}$ which is an infinite subsequence of $\displaystyle \\{p_{n}\pi/2\\}$ and so the series $\displaystyle \sum \frac{1}{n^{k+\sin n}}$ is divergent.

Solution 2:

Here is a handwaving argument that the series diverges for all $k\in(0,2)$: Given $k$, there will be a certain density of $n$ for which $k+cos n <1$. A "random sample" of the harmonic series will diverge by the same argument that shows the harmonic series diverges. Our series will be greater, term by term, than the harmonic series that has numerator 0 if $k+cos n >1$ and numerator 1 if $k+cos n <1$

Solution 3:

If $k$ is a real number less than 2 then $\sum_{n=1}^\infty n^{-k-\sin n}$ diverges.

This is problem 11162 from the American Mathematical Monthly, the solution by the Microsoft Research Problems Group appears in the February 2007 issue.

Here's the solution (if I've transcribed it correctly).

If $k\leq 0$, then the series diverges by comparison with the harmonic series. If $0< k <2$, then let $\alpha,\beta$ be such that $\sin\alpha=\sin\beta=1-k$ and $\pi/2 < \alpha < \beta < 5\pi/2$. Note that $\sin x < 1-k$ for all $x$ with $\alpha < x< \beta$. Define $$ S=(n\in{\mathbb N}:\sin n<1-k )={\mathbb N}\cap\bigcup_{j\in {\mathbb Z}}(2\pi j+\alpha,2\pi j+\beta).$$

Since $\pi$ is irrational, the sequence $n/(2\pi)$ mod 1 is dense in [0,1], so $S\not=\emptyset$ and there exist positive integers $a,A,b,B$ with $$0 < a-2\pi A < {\beta-\alpha\over2}\quad\mbox{and}\quad -{\beta-\alpha\over2} < b-2\pi B < 0.$$

Let $C=\max(a,b)$. Given $s\in S$, we claim that there exists $s^\prime\in S$ with $s < s^\prime\leq s+C$. Indeed, there exists $j\in{\mathbb Z}$ with $\alpha < s-2\pi j<\beta$. Compare $s-2\pi j$ to $(\alpha+\beta)/2$. If $\alpha < s-2\pi j\leq (\alpha+\beta)/2$, then $\alpha< (s+a)-2\pi(j+A)<\beta$, so we can use $s^\prime=s+a$. On the other hand, if $(\alpha+\beta)/2 < s-2\pi j < \beta$, then $a<(s+b)-2\pi(j+B) < \beta$, so we can use $s^\prime=s+b$.

Thus $S$ is infinite. Index it in increasing order, $s_1 < s_2 < \cdots.$ Now $s_j\leq s_1+C(j-1)$, so $$\sum_{n=1}^{s_m}n^{-k-\sin n}\geq \sum_{1\leq n\leq s_m, n\in S} n^{-k-\sin n} \geq \sum_{1\leq n\leq s_m , n\in S} {1\over n} \geq \sum_{j=1}^m {1\over s_j}\geq \sum_{j=1}^m {1\over s_1+C(j-1)},$$ which diverges as $m\to\infty$ by the integral test.

Solution 4:

There is a very interesting paper by Bernard Brighi on the divergence of the series

$\sum\limits_{n=1}^{\infty} \frac{1}{n^{2+ \cos{(a+n)}}}$ for any real $a$ which I would like to attach but I don't know how to do it. Maybe most people who like the series already know it

The link is here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=361997&p=1983209#p1983209