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$\sum_{n=1}^{\infty} u_n$ when $u_1$ is arbitrary real number and $u_{n+1} = \dfrac{1}{n} e^{-u_n}$.

I think that this series diverges. This was my approach:

Since $$\lim_{n\to\infty} u_{n} = \lim_{n\to\infty} u_{n+1}$$

If $\lim_{n\to\infty} u_{n+1} \neq 0$ then it's known that $\sum_{n=1}^{\infty} u_{n}$ diverges.

And if $\lim_{n\to\infty} u_{n+1} = 0$ then I think that somehow I need to show that $u_{n} \geq \dfrac{1}{n}$ and then it's diverges.

How exactly do I prove that?

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    Showing $u_n \leqslant \frac{1}{n}$ won't show divergence. But showing $u_n > \frac{1}{2n}$ will. – 2017-01-10

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Prove, using the mean value theorem, that $e^{-x} > 1 - x$ for all $x > 0$. For all $n \ge 2$, $$u_{n+1} = \frac{e^{-u_n}}{n} >\frac{1-u_n}{n}$$ On the other hand, $u_n < \frac{1}{n-1}$ for all $n \ge 2$. So for all $n \ge 2$,

$$u_{n+1} > \frac{1}{n}\left(1 - \frac{1}{n-1}\right)= \frac{n-2}{n(n-1)}$$

Since $\sum \frac{n-2}{n(n-1)}$ diverges (by the ratio test), then by comparison the series $\sum u_n$ diverges.