$\lim_{n\to \infty}(\lim_{x\to 0}( 1+\sum_{k=1}^n\sin^2(kx))^\frac{1}{n^3x^2} )$

Question

$$\lim_{n\to \infty}\left(\lim_{x\to 0}\left( 1+\sum_{k=1}^n\sin^2(kx)\right)^\frac{1}{n^3x^2} \right)$$

$$=\lim_{n\to \infty}\left(\lim_{x\to 0}\left( 1+\sum_{k=1}^n(k^2x^2)\frac{\sin^2(kx)}{k^2x^2}\right)^\frac{1}{n^3x^2}\right)$$$$=\lim_{n\to \infty}\left(\lim_{x\to 0}\left( 1+\sum_{k=1}^n(k^2x^2)\right)^\frac{1}{n^3x^2}\right)$$$$=\lim_{n\to \infty}\left(\lim_{x\to 0}\left( 1+x^2\sum_{k=1}^nk^2\right)^\frac{1}{n^3x^2}\right)$$$$=\lim_{n\to \infty}\left(\lim_{x\to 0}\left( 1+x^2\left(\frac{n(n+1)(2n+1)}{6}\right)\right)^\frac{1}{n^3x^2}\right)$$$$=\lim_{n\to \infty}e^{\frac{(n+1)(2n+1)}{6n^2}}$$$$=e^\frac13$$

This is how I solved it and the answer is correct according to the answer key provided, yet I am a bit uncomfortable with the solution. That is because when I multiply and divide $\sin^2(kx)$ by $k^2x^2$, I know that $\frac{\sin^2(kx)}{k^2x^2}$ will equal to $1$ if $k^2x^2$ tends to $0$, which is true for all finite values of $k$ where $x \to 0$. However, the expression I get from it is being used to calculate the limit when $k \to n$ where $n \to \infty$, but, as I said, $\frac{\sin^2(kx)}{k^2x^2}$ will equal to $1$ if $k^2x^2$ tends to $0$, which is true for all finite values of $k$. So how is this valid?

Answer 1

As mentioned in the comments, you transform $\frac{\sin u}{u}\overset{?}{=}1$ inside the limit which may or may not lead to a correct answer. Because the rate at which $\sin u/u\to1$ matters and it may be that other factors in the limit, e.g. the exponentiation to the power $1/n^3x^2$, changes the behaviour. As a very basic counterexample, consider: $$\lim_{u\to0}\frac{\sin u-u}{u^3}$$You might be tempted to do this: $$\lim_{u\to0}\frac{\sin u-u}{u^3}=\lim_{u\to0}\frac{\frac{\sin u}{u}-1}{u^2}\overset{??}{=}\lim_{u\to0}\frac{0}{u^2}=0$$But in fact the limit is $-\frac{1}{6}\neq0$.

Fix $n\in\Bbb N$. We can apply L'Hopital's rule twice, or use asymptotic notations: $$\begin{align}\lim_{x\to0}\frac{1}{n^3x^2}\ln\left(1+\sum_{k=1}^n\sin^2(kx)\right)&=\lim_{x\to0}\frac{1}{2n^3x}\cdot\frac{\sum_{k=1}^nk\cdot\sin(2kx)}{1+\sum_{k=1}^n\sin^2(kx)}\\&=\frac{1}{2n^3}\cdot\lim_{x\to0}\frac{\sum_{k=1}^nk\cdot\sin(2kx)}{x}\\&=\frac{1}{n^3}\cdot\lim_{x\to0}\sum_{k=1}^nk^2\cdot\cos(2kx)\\&=\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}\end{align}$$

You may object: it looks like I just treated “$\sin(kx)=0$” there (replacing $1+\sum\sin^2$ with $1$), seemingly contradicting what I said at the start. But here I know for certain it’s ok to do so because there is an easy theorem: if $f(x)\to a$ and $g(x)\to b\neq0$ as $x\to c$ then $f(x)/g(x)\to a/b$ as $x\to c$.

Therefore: $$\left(1+\sum_{k=1}^n\sin^2(kx)\right)^{\frac{1}{n^3x^2}}\underset{x\to0}{\longrightarrow}\exp\left(\frac{n(n+1)(2n+1)}{6n^3}\right)$$For any $n\in\Bbb N$.

So you have: $$\lim_{n\to\infty}\exp\left(\frac{n(n+1)(2n+1)}{6n^3}\right)=\exp\left(\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}\right)=\exp(1/3)$$

Answer 2

Just for the fun of finding the next term.

$$A=\sum_{k=1}^n \sin^2(kx)=\frac{1}{4} (2n+1-\csc (x)\, \sin ((2 n +1)x))$$ Expanding around $x=0$ $$A=\frac{1}{6} n (n+1) (2 n+1) x^2-\frac{1}{90}n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)x^4+O(x^6)$$ Adding $1$ and taking the logarithm $$\log(1+A)=\frac{1}{6} n (n+1) (2 n+1) x^2-$$ $$\frac{1}{360} n (n+1) (2 n+1) \left(10 n^3+27 n^2+17 n-4\right)x^4+O(x^6)$$ $$B=\frac{\log(1+A) } {n^3\,x^2}=\frac{(n+1) (2 n+1)}{6 n^2}-$$ $$\frac{1}{360 n^2} (n+1) (2 n+1) \left(10 n^3+27 n^2+17 n-4\right)x^2+O(x^4)$$ $$e^B=e^{\frac{(n+1) (2 n+1)}{6 n^2}} \left(1-\frac{(n+1) (2 n+1) \left(10 n^3+27 n^2+17 n-4\right)}{360 n^2} x^2+O(x^4) \right)$$

Then the limit and a bit more.