Trying to evaluate a limit of a sum of cosines and am stuck on something that looks like a Riemann sum,

Question

here's where I am stuck:

$$\lim_{n \to \infty} \sum_{k=0}^{n-1} \cos(kx/n)\frac{1}{n}$$

so...it looks like at this point I could convert to a Riemann integral, but to which one?

Maybe $$\int_0^{\infty} \cos(t) dt?$$

That wouldn't converge, so it wouldn't be good.

How could I proceed? It's easy to accept that $\large \frac{1}{n}$ maps to $dt$, when going from a Riemann sum to a Riemann integral, but I particularly have trouble figuring out how to convert the discrete variable to the continuous, integration variable, t.

Any ideas are welcome.

Thanks,

Answer 1

Hint: How would you write $$\int_0^1 \cos(xt) \,dt$$ as a Riemann sum?

Answer 2

Consider the function $f(t) = \cos(tx)$ on $[0,1]$. Let

$$ 0 = t_0 < t_1 = \frac{1}{n} < t_2 = \frac{2}{n} < \ldots < t_{n-1} = \frac{n-1}{n} < t_n = 1$$ be a partition of $[0,1]$ into $n$ sub-intervals of length $\frac{1}{n}$. In the interval $[t_i, t_{i+1}]$ choose the point $\xi_i = t_i$. Then the Riemann sum corresponding to the partition $\{t_i\}_{i=0}^{n-1}$ and the choice of points $\{\xi_i\}_{i=0}^{n-1}$ is

$$ \sum_{k=0}^{n-1} f(\xi_k) (t_{k+1} - t_k) = \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \frac{1}{n} = \sum_{k=0}^{n-1} \cos \left( \frac{k}{n} x \right) \frac{1}{n}. $$

Hence, the limit of the expression (as $n \to \infty$) converges to

$$ \int_{0}^1 \cos(tx) \, dt = \begin{cases} [\frac{\sin(tx)}{x}]^{t = 1}_{t=0} & x \neq 0, \\ 1 & x = 0. \end{cases} = \begin{cases} \frac{\sin(x)}{x} & x \neq 0, \\ 1 & x = 0. \end{cases} $$