Trigonometry limit's proof: $\lim_{x\to0}\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}=\frac{k(k+1)}{2}$

Solution 1:

You are so close. Note that \begin{align*} \frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x} &= \frac{\sin(x)}{x}+\frac{\sin(2x)}{x}+\cdots+\frac{\sin(kx)}{x} \\ &= \frac{\sin(x)}{x}+2\frac{\sin(2x)}{2x}+\cdots+k\frac{\sin(kx)}{kx} \\ &\to 1 + 2 + \cdots + k \\ &= \frac{k(k+1)}{2} \end{align*} as $x\to0$.

I suspect you may not have been able to finish because you didn't recognize the identity $$ 1 + 2 + \cdots + k = \frac{k(k+1)}{2}. $$ This identity has a very cute proof. Set $S:=1+2+\cdots+k$. Adding \begin{align*} 1 + 2 + \cdots + k &= S \\ k + (k-1) + \cdots + 1 &= S \\ \end{align*} gives \begin{align*} \underbrace{(k+1)+(k+1)+\cdots+(k+1)}_{k\ \text{times}} = 2S. \\ \end{align*} Therefore $k(k+1)=2S$ and consequently $$1+2+\cdots+k = S = \frac{k(k+1)}{2}.$$

Solution 2:

This is the derivative at $0$ of the function $$ f(x)=\sin(x)+\sin(2x)+\dots+\sin(kx) $$ and $$ f'(x)=\cos(x)+2\cos(2x)+\dots+k\cos(kx) $$ Therefore $$ f'(0)=1+2+\dots+k $$ and you're done with child Gauss’ trick.

^{Note: This isn't using l’Hôpital, but the definition of derivative and the chain rule.}