2
$\begingroup$

Why doesn't the following limit exist? $$\lim_{n\rightarrow \infty }\left(1+\frac{\cos(n\pi )}{n}\right)^{n}$$

I write that because $\cos$ has values in $[-1,1]$ and $n\rightarrow \infty$ this limit can't exists but I don't know if I'm right.

$\endgroup$
  • $\begingroup$ The expression amounts to $\left(1\pm\frac1n\right)^n$ which tends to $e^{\pm1}$. $\endgroup$ – Yves Daoust Feb 2 at 16:12
4
$\begingroup$

$\cos (n \pi)=(-1)^n $. Now look at the subsequences $(a_{2n}) $ and $(a_{2n+1}) $.

$\endgroup$
  • $\begingroup$ Thank you for your help :) $\endgroup$ – DaniVaja Feb 2 at 16:39
  • $\begingroup$ So limit of $a_{2n}=(1+\frac{cos(2n\pi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+\frac{cos(2n\pi+\pi )}{2n+1})^{2n+1}$ because $\cos (n \pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct ? $\endgroup$ – DaniVaja Feb 2 at 16:43
4
$\begingroup$

Hint:

A sequence $a_n$ convergence to $L$ if and only if every sub-sequence convergence to $L$.

Consider the sub-sequences $a_{2n},a_{2n+1}$ of $a_n =(1+\frac{\cos(n\pi )}{n})^{n}$.

$\endgroup$
  • $\begingroup$ So limit of $a_{2n}=(1+\frac{cos(2n\pi )}{2n})^{2n}$ is different from limit of $a_{2n}=(1+\frac{cos(2n\pi+\pi )}{2n+1})^{2n+1}$ because $\cos (n \pi)=(-1)^n$ which gives me $1$ or $-1$ and that's why limit doesn't exists, because cos oscilating between $-1$ and $1$.Is my answer correct? $\endgroup$ – DaniVaja Feb 2 at 16:39
  • 1
    $\begingroup$ @DaniVaja yes that's exactly the idea. $\endgroup$ – Yanko Feb 3 at 11:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.