Intereting Posts

Convergence of the series $\sum\limits_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}$
Distribution of the lifetime of a system consisting of two exponentially distributed components, one being backup
Compute the mean of $(1 + x)^{-1}$
Riemann Zeta Function Manipulation
What exactly are eigen-things?
Some questions about Banach's proof of the existence of continuous nowhere differentiable functions
Bound for the degree
Uniform Continuity implies Continuity
Integral change that I don't understand
Separable implies second countable
on the boundary of analytic functions
Digonalizable Polynomial Matrix
A question about the vector space spanned by shifts of a given function
The security guard problem
A Banach space is reflexive if and only if its dual is reflexive

We know that the “Harmonic Series” $$ \sum \frac{1}{n}$$ diverges. And for $p >1$ we have the result that the series converges $$\sum \frac{1}{n^{p}}$$ converges.

One can then ask the question of testing the convergence the following 2 Series:

$$\sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \cos{n}}}, \quad \sum\limits_{n=1}^{\infty} \frac{1}{n^{k + \sin{n}}}$$ where $ k \in (0,2)$.

- functions with eventually-constant first difference
- Show that $\sum_{n = 1}^\infty n^qx^n$ is absolutely convergent, and that $\lim_{n \rightarrow \infty}$ $n^qx^n = 0$
- Isometric Embedding of a separable Banach Space into $\ell^{\infty}$
- Can I exchange limit and differentiation for a sequence of smooth functions?
- Continuity and the Axiom of Choice
- Proving the limit at $\infty$ of the derivative $f'$ is $0$ if it and the limit of the function $f$ exist.

Only thing which i have as tool for this problem is the inequality $| \sin{n} | \leq 1$, which i am not sure whether would applicable or not.

- How to prove a sequence does not converge?
- A function continuous on all irrational points
- Two exercises on characters on Marcus (part 1)
- Are Continuous Functions Always Differentiable?
- Proof that every polynomial of odd degree has one real root
- Proving that $(\forall\epsilon>0)(\exists n,k\in\mathbb{N})(|\frac{n}{k}-\pi|<\frac{\epsilon}{k})$
- Real Analysis - Prove limit $\lim_{x\to 25} \sqrt x = 5$
- Is there an integral for $\pi^4-\frac{2143}{22}$?
- Young's inequality without using convexity
- real analysis function takes on each value twice?

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.

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 $k+cos n <1$

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.

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

- Question about Cardinality
- Green's theorem exercise
- Can locally “a.e. constant” function on a connected subset $U$ of $\mathbb{R}^n$ be constant a.e. in $U$?
- ln(z) as antiderivative of 1/z
- Combinatorics question about choosing non consecutive integers
- Understanding differentials
- Cardinality of the quotient ring $\mathbb{Z}/(x^2-3,2x+4)$
- When is a metric space Euclidean, without referring to $\mathbb R^n$?
- Some questions about Hartshorne chapter 2 proposition 2.6
- Proving Inequality using Induction $a^n-b^n \leq na^{n-1}(a-b)$
- Unique manifold structure
- How would one be able to prove mathematically that $1+1 = 2$?
- How to prove that the problem cannot be solved by the four Arithmetic Operations?
- Field in $\mathbb{F}_3$
- A coin is tossed $m+n$ times $(m>n)$.Show that the probability of atleast $m$ consecutive heads is $\frac{n+2}{2^{m+1}}$