I'm facing a bit of trouble figuring out this limit.

$$ \lim_{n \to \infty} \cos\left(\left(-1\right)^n \frac{n-1}{n+1}\pi\right)$$

and I'm not sure if I can simply find the limit of the inner functions and then apply cosine to that, as in $$ \lim_{n \to \infty} (-1)^n = undefined \quad \quad \lim_{n \to \infty} \frac{n-1}{n+1} = 1 \quad \quad \lim_{n \to \infty} \pi = \pi $$ But because of the oscillation caused by $\displaystyle\lim_{n \to \infty} (-1)^n$, I am not sure what I should do. It would seem to me that the entire thing is undefined, but that is a bad answer.

  • $\begingroup$ $n\to\infty$ does not necessarily imply $n$ is an integer, for non integer $n,(-1)^n=?$ $\endgroup$ – lab bhattacharjee Apr 28 '18 at 18:11
  • $\begingroup$ True. Then it would tend to oscillate. I'm not quite sure what I can do with that then. Thanks $\endgroup$ – user25758 Apr 28 '18 at 18:20
  • $\begingroup$ Usually $n$ is used to denote an integer, so assuming it is, note that $\cos(-\pi)=\cos(\pi)=-1$. $\endgroup$ – Cedron Dawg Apr 28 '18 at 19:06

Note that $\cos$ is even; that is, $\cos(x)=\cos(-x)$ for any $x\in\mathbb{R}$. Thus,

  • If $n$ is odd, then $\cos\left((-1)^n \frac{n-1}{n+1}\pi\right)=\cos\left((-1)^{n+1}\frac{n-1}{n+1}\pi\right)=\cos\left(\frac{n-1}{n+1}\pi\right)$.
  • If $n$ is even, then $\cos\left((-1)^n \frac{n-1}{n+1}\pi\right)=\cos\left(\frac{n-1}{n+1}\pi\right)$.

Therefore, $$\lim_{n\to\infty}\cos\left((-1)^n \frac{n-1}{n+1}\pi\right)=\lim_{n\to\infty}\cos\left(\frac{n-1}{n+1}\pi\right)=\cos\pi=-1.$$

  • $\begingroup$ Simple and clear explanation +1 $\endgroup$ – Paramanand Singh Dec 22 '18 at 7:33

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