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

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    $\begingroup$ You should write $\cos^{2n}(x-10)$ or $[\cos(x-10)]^{2n}.$ $\endgroup$ – zhw. May 26 at 15:59
  • $\begingroup$ @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 $n=10$. $\endgroup$ – Vineet Mangal May 26 at 17:47
  • $\begingroup$ @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. $\endgroup$ – Julian Mejia May 27 at 1:38
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This is a finite sum, just show that $\cos(x-10)^{2n}\to 0$ except for $x=10$, where you get $1$.

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  • $\begingroup$ Actually, except for $x=10+n\pi,n\in \mathbb Z.$ $\endgroup$ – zhw. May 26 at 16:00
  • $\begingroup$ @zhw, sure but I was referring for the $x\in \{1,2,\dots,20\}$. $\endgroup$ – Julian Mejia May 26 at 16:12
  • $\begingroup$ Ah, I didn't read the problem properly. Thanks. $\endgroup$ – zhw. May 26 at 16:23

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