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    #1
    $\displaystyle s_n=\left(\frac1n-1\right)^n$

    My attempt:

    For large $n$, the sequence oscillates between $e^{-1}$ and $-e^{-1}$ and therefore diverges. Now for the proof.

    Assume, for the sake of argument, that the sequence converges to $L$.

    $\exists N\in\mathbb{N}$ such that $|s_n-L|<0.1$ whenever $n\ge N$

    $\displaystyle\left|\left(\frac1n-1\right)^n-L\right|<0.1$ whenever $n\ge N$

    $\displaystyle\implies\left|\left(\frac1{n+1}-1\right)^{n+1}-L\right|<0.1$ whenever $n\ge N$

    We can rewrite these 2 equations as

    $\displaystyle\left|(-1)^n\left(1-\frac1n\right)^n-L\right|<0.1$ whenever $n\ge N$ --------------- (1)

    $\displaystyle\left|(-1)^{n+1}\left(1-\frac1{n+1}\right)^{n+1}-L\right|<0.1$ whenever $n\ge N$

    $\displaystyle\implies\left|(-1)^n\left(1-\frac1{n+1}\right)^{n+1}+L\right|<0.1$ whenever $n\ge N$ --------------- (2)

    How do I get a contradiction from equations (1) and (2)?

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    #2
    Quote Originally Posted by Alexmahone View Post
    $\displaystyle\left|(-1)^n\left(1-\frac1n\right)^n-L\right|<0.1$ whenever $n\ge N$ --------------- (1)

    $\displaystyle\left|(-1)^{n+1}\left(1-\frac1{n+1}\right)^{n+1}-L\right|<0.1$ whenever $n\ge N$

    $\displaystyle\implies\left|(-1)^n\left(1-\frac1{n+1}\right)^{n+1}+L\right|<0.1$ whenever $n\ge N$ --------------- (2)

    How do I get a contradiction from equations (1) and (2)?
    If it is known that $(1-1/n)^n\to1/e$, then you can reason as follows. Suppose $|(1-1/n)^n-1/e|\le0.1$ for all $n\ge N$ (we can achieve this increasing $N$ if necessary). Then (1) with an even $n$ implies that $|L-1/e|<0.2$ and therefore $L>1/e-0.2>0$ and (2) implies that $|(-L)-1/e|<0.2$ and therefore $L<-1/e+0.2<0$, which is a contradiction.

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