Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges

Question

How to prove that $$u_n=\sum_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$$ is a Cauchy sequence?

The exercise I am reading gives as a hint that we should use the inequality: $$0\leq\cos\left(\frac kn\right)^{2n^2/k}\leq e^{-k},$$ for all $k\leq n$.

I tried to estimate $|u_m-u_n|$, but I don't know how to deal with the $n$ inside the sum.

Answer 1

Forget Cauchy sequences. To show the convergence of the sequence $(u_n)$, consider, for every integer positive $(k,n)$, $$v_{k,n}=\cos\left(\frac{k}{n}\right)^{2n^2/k}\mathbf 1_{k\leqslant n}$$ then $$u_n=\sum_{k=1}^\infty v_{k,n}$$ Assume momentarily that, for every $x$ in $[0,1]$, $$\cos x\leqslant e^{-x^2/2}\tag{$\ast$}$$ Then, for every positive $(k,n)$, $$|v_{k,n}|=v_{k,n}\leqslant\left(e^{-k^2/(2n^2)}\right)^{2n^2/k}=e^{-k}$$ and, for every fixed positive $k$, $$\lim_{n\to\infty}v_{k,n}=e^{-k}$$ Thus, by Lebesgue dominated convergence theorem for series, $$\lim_{n\to\infty} u_n=\sum_{k=1}^\infty\lim_{n\to\infty} v_{k,n}=\sum_{k=1}^\infty e^{-k}=\frac1{e-1}$$ To complete the proof, one needs to prove $(\ast)$ but this is direct, considering the derivative of the function $$f(x)=e^{x^2/2}\cos x$$ on $[0,1]$ and using the estimate $\tan x\geqslant x$ on this interval.

Answer 2

Hint. You have at least two options:

• use formulas for geometric series,
• use a comparison with the integral $\int_{-\infty}^0e^{t}dt=1$

Answer 3

First of all, $\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}\leq e^{-k} $ is equivalent to $\cos\left(\frac{k}{n}\right) \leq e^{-k^2/2n^2} $ which follows from $e^{-x} \ge 1-x+x^2/2$ for $\frac12 \ge x \ge 0$ (so $e^{-k^2/2n^2} \ge 1-k^2/(4n^2)+k^4/(8n^4) $) and $\cos(x) \le 1-x^2/2+x^4/24$ (so $\cos(k/n) \le 1-k^2/(2n^2)+k^4/(24n^4) $). Note that we have to go to the $x^4$ term.

Then $u_n =\sum_{k=1}^n\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}} \le\sum_{k=1}^n e^{-k} \lt \frac{1}{e-1} $ so $u_n$ is a bounded sequence.

However, we have not yet shown that $u_n$ is increasing.

Instead I will show that $u_n \to \frac{1}{e-1} $ with too much computation.

We have $\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}} \ge \left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k}} = \left(\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}\right)^k $.

For $0 < x \le \frac12$,

$\begin{array}\\ -\ln(1-x) &=\sum_{k=1}^{\infty} \dfrac{x^k}{k}\\ &=x+\sum_{k=2}^{\infty} \dfrac{x^k}{k}\\ &\lt x+\sum_{k=2}^{\infty} \dfrac{x^k}{2}\\ &=x+\dfrac{x^2}{2(1-x)}\\ &\le x+x^2/4\\ \end{array} $

or $\ln(1-x) \ge -x-x^2/4 $.

Therefore, for $0 < x \le \frac12$, $(1-x)^{1/x} =\exp((1/x)\ln(1-x)) \ge\exp(-(1/x)(x+x^2/4)) =\exp(-1-x/4) $ so $\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}} \ge \exp(-1-k^2/(8n^2)) $ so $\left(\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}\right)^k \ge \exp(-k-k^3/(8n^2)) $ so $u_n =\sum_{k=1}^n\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}} \ge\sum_{k=1}^n \exp(-k)\exp(-k^3/(8n^2)) $.

I will now split the sum into two parts: $[1, n^c]$ and $(n^c, n]$ where $0 < c < 1$.

If $k > n^{c}$, all the terms are positive, so that sum is positive.

For the rest of $u_n$, since $e^{-x} \gt 1-x $,

$\begin{array}\\ \sum_{k=1}^{n^{c}} \exp(-k)\exp(-k^3/(8n^2)) &\gt \sum_{k=1}^{n^{c}} \exp(-k)(1-k^3/(8n^2))\\ &= \sum_{k=1}^{n^{c}} \exp(-k)-\sum_{k=1}^{n^{c}} \exp(-k)k^3/(8n^2)\\ &= \sum_{k=1}^{n^{c}} \exp(-k)-\sum_{k=1}^{n^{c}} \exp(-k)n^{3c-2}/8\\ &\gt \sum_{k=1}^{n^{c}} \exp(-k)-\frac{n^{3c-2}}{8}\sum_{k=1}^{n^{c}} \exp(-k)\\ \end{array} $

and the first term approaches $\sum_{k=1}^{\infty} \exp(-k) =\frac{1/e}{1-1/e} =\frac{1}{e-1} $ and the second term is less than $\frac{n^{3c-2}}{8(e-1)} $ which goes to zero if $c < \frac23$.

Therefore the sum approaches $\frac{1}{e-1}$.

Whew.

That was harder than I thought it would be.

Hope that it's correct.

Answer 4

Starting from the high school limit $\lim_{x\to 0}\log(\cos(x))/x^2=-1/2$, we have near $0$ that $\cos(x)\approx e^{-x^2/2}$. Intuitively (having in mind the Taylor series and focusing on the interval $x$ in $[0,1]$), it's clear now you can bound the sum using the exponential function as you state in your post.

More explanations: The high school limit gives us the starting idea, and the Taylor series of $e^{-x^2/2}-\cos(x)$, with $x$ in $[0,1]$ assures the desired inequality.

Answer 5

By making use of the inequality $0\leq cos(\frac{k}{n})^{2n^2/k}\leq e^{-k}$ for all $k\leq n$ we can state that

$$\sum_{k=1}^n cos(\frac{k}{n})^{2n^2/k}\leq\sum_{k=1}^ne^{-k}$$

If an $f$ function verifies that $f'(x)<0$ and $f''(x)>0$ for all $x\in[u,v]$ then we can apply the following inequality

$$\sum_{k=u}^vf(k)\leq\int_u^{v+1}f(x)dx+f(u)-f(v+1)$$

As the function $e^{-x}$ meets the requirements of the statement for all $x\in\mathbb{R}$ we can prove

$$\sum_{k=1}^ne^{-k}\leq\int_1^{n+1}e^{-x}dx+e^{-1}-e^{-n-1}$$

If $n\in\mathbb{N}$ it is convergent, so we just have to know what happens when $n\rightarrow\infty$. Hence,

$$\lim_{n\rightarrow\infty}\sum_{k=1}^ne^{-k}\leq\lim_{n\rightarrow\infty} \int_1^{n+1}e^{-x}dx+e^{-1}-e^{-n-1}$$

$$\lim_{n\rightarrow\infty}\sum_{k=1}^ne^{-k}\leq -e^{-x}|_1^\infty+e^{-1}=\frac{2}{e}$$