Solve $\lim_{N\to\infty} \frac{\sum_{n=0}^N \sin^4 (5n)}{N}$

Question

$$\lim_{N\to\infty} \frac{\sum_{n=0}^N \sin^4 (5n)}{N}$$

This is an Olympiad problem from India. I tried solving it by trying to manipulate such that n/N is x and then integrate it along with reducing the sin^4 to cos using double angle formulae twice. But I got stuck when a term to be integrated resembled cosx/x which I was unable to integrate by the methods I know. But judging the kind of problems this exam asks, I think there is a more elegant approach that does not require knowledge of integrating the aforementioned.

Source

Answer 1

Consider $5n$ mod $\pi$. As $n$ increases, you get a dense covering of $[0,\pi]$. So one way to express the following integral as a Riemann sum is $$\int_0^{\pi}\sin^4(x)\,dx=\lim_{N\to\infty}\sum_{n=1}^N\sin^4(5n)\Delta x_n$$ You could argue the sampling is uniformly dense such that it is OK to take $\Delta x_n=\frac{\pi}{N}$. Then you conclude $$\begin{align} \lim_{N\to\infty} \frac{\sum_{n=0}^N \sin^4 (5n)}{N} &=\lim_{N\to\infty} \frac{\sum_{n=1}^N \sin^4 (5n)}{N}\\ &=\frac{1}{\pi}\int_0^{\pi}\sin^4(x)\,dx\\ &=\frac{1}{4\pi}\int_0^{\pi}\left(1-\cos(2x)\right)^2\,dx\\ &=\text{... more elementary integration ...}\\ &=\frac{1}{4\pi}\int_0^{\pi}\frac32\,dx\\ &=\frac38 \end{align}$$

Answer 2

\begin{align}&\frac1N\sum_{k=0}^N\sin^4(5k)=\\=&\frac1N\sum_{k=0}^N\left(\frac{e^{5ik}-e^{-5ik}}{2i}\right)^4=\frac1{16N}\sum_{k=0}^Ne^{20ik}-4e^{10ik}+6-4e^{-10ik}+e^{-20ik}=\\=&\frac1{16N}\left(\frac{e^{20i(N+1)}-1}{e^{20i}-1}+\frac{e^{-20i(N+1)}-1}{e^{-20i}-1}-4\frac{e^{4i(N+1)}-1}{e^{4i}-1}-4\frac{e^{-4i(N+1)}-1}{e^{-4i}-1}+6(N+1)\right)=\\=&\frac38+\frac1{16N}\left(\frac{e^{20i(N+1)}-1}{e^{20i}-1}+\frac{e^{-20i(N+1)}-1}{e^{-20i}-1}-4\frac{e^{4i(N+1)}-1}{e^{4i}-1}-4\frac{e^{-4i(N+1)}-1}{e^{-4i}-1}+1\right)\end{align}

And here it depends how familiar are with limits of complex-valued functions. I would say that the quantity between parenthesis is bounded in $N$ (for instance, its absolute value is bounded by $1+\frac{8}{(\cos20-1)^2+\sin^220}+\frac{32}{(\cos 4-1)^2+\sin^24}$) and therefore the limit as $N\to \infty$ is $\frac38$. Otherwise you calculate the $a+bi$ form of that quantity (which will have $b=0$) and then draw the same conclusion with the resulting expression in $\sin(N+1)$ and $\cos(N+1)$.

Answer 3

@Gae.S has answered before me.

Here's an approach using trigonometry

$\begin{align}\sin^4(5n) = \left(\dfrac{1-\cos(10n)}{2}\right)^2 = \dfrac{1-2\cos(10n)+\cos^2(10n)}{4} = \dfrac{1-2\cos(10n)+\frac{1+\cos(20n)}{2}}{4}=\dfrac{3-4\cos(10n)+\cos(20n)}{8}\end{align}$

Now $\begin{align}\lim_{n\to\infty}\frac1N\sum_{n=0}^N\frac38 = \lim_{N\to\infty}\frac38\left(\frac{N+1}{N}\right)=\frac38\end{align}$

Using the formula,

$$\sum_{n=0}^N\cos(kn) = \frac{\sin((N+1)\frac k2)}{\sin\frac k2}\cos\frac {Nk}2$$

$$\begin{align}&\lim_{N\to\infty}\frac1N\sum_{n=0}^N\frac{\cos(20n)-4\cos(10n)}{8} \\&= \lim_{N\to\infty}\frac{\sin(10N+10)}{8\sin(10)}\frac{\cos(10N)}N-\lim_{N\to\infty}\frac{\sin(5N+5)\cos(5N)}{2\sin(5)}\frac{\cos(5N)}N\\ &=0 ~~(\text{As}, \lim_{x\to\infty}\frac{\cos x}x = 0)\end{align}$$

So the final result is $\frac38$