# Showing that $\lim_{n \to \infty} \sum_{x = 1}^{20} \cos(x-10)^{2n} = 1\$

How do I show that $$\lim_{n \to \infty} \sum_{x = 1}^{20} \cos(x-10)^{2n} = 1\ ?$$

• You should write $\cos^{2n}(x-10)$ or $[\cos(x-10)]^{2n}.$ – zhw. May 26 at 15:59
• @zhw. Well, I think he wrote that correct because if the power $2n$ is on $cos$, then sum can't be zero for $n= 1$ to $20$ except for $n=10$. – Vineet Mangal May 26 at 17:47
• @Vineet Mangal, the remark of zhw is correct(and I considered this in my solution), if the term were $\cos [(x-10)^{2n}]$ the statement wouldn't be true. – Julian Mejia May 27 at 1:38

This is a finite sum, just show that $$\cos(x-10)^{2n}\to 0$$ except for $$x=10$$, where you get $$1$$.
• Actually, except for $x=10+n\pi,n\in \mathbb Z.$ – zhw. May 26 at 16:00
• @zhw, sure but I was referring for the $x\in \{1,2,\dots,20\}$. – Julian Mejia May 26 at 16:12