Why $\sum_{k=1}^m \left(\cos{\frac{2\pi k}{2m+1}}\right)^{(2m+1)^2}$ converges to $\sum_{k=1}^\infty e^{-2\pi^2k^2}$?

Question

While I was considering random walk on $S^1$, I needed to compute the following.

$$\lim_{m\rightarrow\infty}\sum_{k=1}^m \left(\cos{\frac{2\pi k}{2m+1}}\right)^{(2m+1)^2}$$

I guessed it should converges to trace of heat kernel on $S^1$. So I tried to use Mathematica. Mathematica says, it converges to following series:

$$\sum_{k=1}^\infty e^{-2\pi^2k^2}$$

But how can we prove this is true?

of course, proving $$\lim_{m\rightarrow\infty}\left(\cos{\frac{2\pi k}{2m+1}}\right)^{(2m+1)^2}=e^{-2\pi^2k^2}$$ is easy for each fixed $k$ (one can apply l'hospital rule or use taylor expansion).

But I don't know how to show $$\lim_{m\rightarrow\infty}\sum_{k=1}^m \left(\cos{\frac{2\pi k}{2m+1}}\right)^{(2m+1)^2}=\sum_{k=1}^\infty e^{-2\pi^2k^2}$$.

-Generalized version of above question is the following one: suppose a doubly indexed sequence $\{a_{mk}\}_{m\in\mathbb{N},1\leq k\leq m}$ and a sequence $\{b_k\}_{k=1}^\infty$ satisfy following two conditions:

(1) $\lim_{m\rightarrow\infty}a_{mk}=b_k$ for each positive integer $k$.

(2) $\sum_{k=1}^\infty b_k$ converges.

Then, can we conclude $\lim_{m\rightarrow\infty}\sum_{k=1}^m a_{mk}=\sum_{k=1}^\infty b_k$? In general it is not true (consider $a_{mk}=b_k+m^{-1/2}$). To make the limit true, what kinds of additional assumptions do we need? Maybe there is some known result about this general question?

Answer 1

A sketch:

By exploiting symmetry of the sum, we can rewrite it as

$$ \begin{align} &\sum_{k=1}^m \left(\cos{\frac{2\pi k}{2m+1}}\right)^{(2m+1)^2} \\ &\qquad = \sum_{k=1}^{m/2} \left(\cos{\frac{\pi 2k}{2m+1}}\right)^{(2m+1)^2} - \sum_{k=1}^{m/2} \left(\cos{\frac{\pi (2k-1)}{2m+1}}\right)^{(2m+1)^2}, \tag{$*$} \end{align} $$

assuming $m$ is even. For fixed $k$ we have

$$ \lim_{m \to \infty} \left(\cos{\frac{\pi 2k}{2m+1}}\right)^{(2m+1)^2} = e^{-(\pi 2k)^2/2} $$

and

$$ \lim_{m \to \infty} \left(\cos{\frac{\pi(2k-1)}{2m+1}}\right)^{(2m+1)^2} = e^{-(\pi(2k-1))^2/2}. $$

Then, applying the dominated convergence theorem to each sum in $(*)$ using the domination relationships

$$ \left(\cos{\frac{\pi 2k}{2m+1}}\right)^{(2m+1)^2} \leq e^{-(\pi 2k)^2/2} $$

and

$$ \left(\cos{\frac{\pi(2k-1)}{2m+1}}\right)^{(2m+1)^2} \leq e^{-(\pi(2k-1))^2/2} $$

yields

$$ \begin{align} \lim_{m \to \infty} \sum_{k=1}^m \left(\cos{\frac{2\pi k}{2m+1}}\right)^{(2m+1)^2} &= \sum_{k=1}^{\infty} \left(e^{-(\pi 2k)^2/2} - e^{-(\pi(2k-1))^2/2}\right) \\ &= \sum_{k=1}^{\infty} (-1)^k e^{-(\pi k)^2/2}. \end{align} $$

Mathematica can evaluate this sum in terms of elliptic functions, and its numerical value is

$$ \sum_{k=1}^{\infty} (-1)^k e^{-(\pi k)^2/2} \doteq -0.00719\ 18806\ 80538\ 37459. $$

In comparison, the Mathematica command

m=200;Sum[Cos[2 Pi k/(2 m+1)]^((2 m+1)^2),{k,1,m}]//N

returns the numerical approximation of the original sum when $m = 200$

$$ \left. \sum_{k=1}^m \left(\cos{\frac{2\pi k}{2m+1}}\right)^{(2m+1)^2} \right|_{m = 200} \doteq -0.00719\ 152. $$

Answer 2

In response to the general case:

Claim:

If $|f_m(k)|\le g(x)$ where $g$ is an integrable function, and $\lim_{m\to\infty}f_m(x)=f(x)$ then $$\lim_{m\to\infty}\sum^m_{k=1}f_m(k)=\sum^\infty_{k=1}f(k)$$

Proof:

Define a ‘step function’: $$ H_t(x)= \begin{cases} 1, & x\le t\\ 0, & x>t \end{cases} $$

Then, $$S(m):=\sum^m_{k=1}f_m(k)=\sum^\infty_{k=1}f_m(x)H_m(x)$$

Let $\mu$ be the counting measure on $\mathbb N$.

Then, $$S(m)=\int f_m(x)H_m(x) d\mu$$

Since $|f_m(x)H_m(x)|\le |f_m(x)|\le g(x)$, by dominated convergence theorem, $$\lim_{m\to\infty}S(m)=\int\lim_{m\to\infty}f_m(x)H_m(x)d\mu=\int f(x) d\mu=\sum^\infty_{k=1}f(k)$$

Q.E.D.

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