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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.

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  • $\begingroup$ The expression amounts to $\left(1\pm\frac1n\right)^n$ which tends to $e^{\pm1}$. $\endgroup$
    – user65203
    Commented Feb 2, 2019 at 16:12

2 Answers 2

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

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  • $\begingroup$ Thank you for your help :) $\endgroup$
    – DaniVaja
    Commented Feb 2, 2019 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
    Commented Feb 2, 2019 at 16:43
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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 $a_n =(1+\frac{\cos(n\pi )}{n})^{n}$.

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  • $\begingroup$ 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
    Commented Feb 2, 2019 at 16:39
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    $\begingroup$ @DaniVaja yes that's exactly the idea. $\endgroup$
    – Yanko
    Commented Feb 3, 2019 at 11:02

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