Converges given $e^{a_n} = a_n + e^{b_n}$ Let $\{ a_n \}$ and $\{ b_n \}$ be two sequences such that for each $n$ we have
$$e^{a_n} = a_n + e^{b_n}$$
If $a_n > 0$ for all $n$ and if $\sum a_n$ converges, show that $\sum \left( \frac{b_n}{a_n} \right)$ converges
 A: Since $\sum_{n\geq 1}a_n$ is convergent, $\lim_{n\to +\infty}a_n = 0$.
So, if $n$ is big enough, $e^{a_n}$ equals $1+a_n+C_n a_n^2$ where $C_n$ is a bounded and close to $\frac{1}{2}$. That gives $e^{b_n}=1+C_n a_n^2$, so $b_n$ behaves like $a_n^2$ and $\sum_{n\geq 1}\frac{a_n}{b_n}$ behaves like $\sum_{n\geq 1}\frac{1}{a_n}$, than cannot be convergent, since its general term is not bounded.
As a further disproof, we may consider $a_n=\frac{1}{n^2}$, giving: 
$$b_n=\log\left(e^{1/n^2}-\frac{1}{n^2}\right)=\frac{1}{2n^4}+O\left(\frac{1}{n^6}\right)$$
as well as:
$$ \frac{a_n}{b_n} = 2n^2 + O(1).$$
On the other hand, the same argument gives that $\frac{b_n}{a_n}$ behaves like $a_n$, hence $\sum_{n\geq 1}\frac{b_n}{a_n}$ is convergent. We also have an almost explicit inequality: since $\frac{\log(e^x-x)}{x}\leq\frac{17 x}{30}$ over $(0,1]$,
$$ \sum_{n:a_n\leq 1}\frac{b_n}{a_n}\leq \frac{17}{30}\sum_{n:a_n\leq 1}a_n.$$
A: Using the Limit Comparison Test:
First observe that, since the series $\sum_n a_n$ converges, we must have
$$\lim_{n\to+\infty}a_n=0.$$
Now, writing
$$\forall n\geq0,\ b_n=\ln\bigl(\mathrm{e}^{a_n}-a_n\bigr)=\ln\bigl(1+(\mathrm{e}^{a_n}-1-a_n)\bigr),$$
and observing that
$$\mathrm{e}^{a_n}-1-a_n=\frac{a_n^2}2+o(a_n^2)$$
yields
$$b_n=\frac{a_n^2}2+o(a_n^2),$$
hence
$$\frac{b_n/a_n}{a_n}=\frac12+o(1)\longrightarrow\frac12\in\mathbb{R}_+^*.$$
Hence (since $(a_n)_n$ is a sequence of positive terms) we conclude, by the Limit Comparison Test, that the series
$$\sum_n\frac{b_n}{a_n}\qquad\text{and}\qquad\sum_na_n$$
have the same nature, hence the series $\sum_nb_n/a_n$ converges too.
