Why is it possible here to sum equivalent

Question

We have a sequence : $$u_n = \frac{e^{-u_{n-1}}}{n+1}$$

It's easy to prove that : $\lim_{n \to \infty} u_n = 0$ and that $\lim_{n \to \infty} nu_n = 1$ so that : $u_n \sim \frac{1}{n}$.

Then I am asked to find the nature of $ \sum (-1)^n u_n$.

In the correction of the exercise they do the following :

$$\mid u_n \mid = \frac{1-u_{n-1} +O(u_{n-1})}{n+1} = \frac{1}{n} - \frac{1}{n^2} + O(\frac{1}{n^2})$$

I don't understand why is it possible to replace $u_{n-1}$ by the equivalent of $u_{n-1}$ which is $\frac{1}{n-1}$ since it's a sum. Normally in a sum we need to do some tricks in order to replace a sequence by it's equivalent.

When I compute the limit of $ \frac{1-u_{n-1}}{1-1/n}$ it does go to $1$ at infinity. But they are not giving any explanations of why it is possible here ? Is there a rule or something I missed ?

Thank you !

Answer 1

Consider the sequences $a_n=\frac{1}{n+1}$, $b_n=u_n-a_n$.

Then $\sum_n{(-1)^na_n}$ is well-known to be convergent.

Furthermore, $b_n=\frac{1}{n+1}(e^{-u_{n-1}}-1)\sim \frac{-u_{n-1}}{n} \sim -n^{-2}$, thus $\sum_n{b_n(-1)^n}$ is absolutely convergent.