# Prove that if $a_n>0$ and $\sum a_n$ converges then $\sum (\frac {b_n}{a_n})$ converges

Let $\{a_n\},\{b_n\}$ be two sequences such that for each $n$ we have $e^{ a_n }= a_n + e^{b_n }$.

Show that if $a_n>0$ and $\sum a_n$ converges $\implies \sum \left(\frac {b_n}{a_n}\right)$ converges.

Attempt: We have : $e^{ a_n }= a_n + e^{b_n }$

$\implies 1 + a_n + \dfrac {a_n^2}{2!} + \cdots =a_n+ 1+ b_n+ \dfrac {b_n^2}{2!} + \cdots$

I am not sure if this is the way I should have begun.

Can somebody please guide me on how to proceed with this problem.

Thank you very much for your help in this regard.

• Hint: $e^x>1+x$ for $x\neq 0$, so $e^{b_n}=e^{a_n}-a_n>1$. – Wojowu Jan 27 '15 at 11:15

Since $\sum a_n$ is convergent and $a_n>0$, we have that $\lim_{n\to +\infty}a_n = 0$, so we are free to assume that for every $n\geq N$, $a_n\leq 1$ holds. Since over the interval $(0,1]$ we have
$$x+\exp(x^2)> \exp(x), \tag{1}$$ it follows that for every $n\geq N$ we have $\color{red}{b_n\leq a_n^2}$, since by assuming $b_n>a_n^2$ we have $e^{b_n}+a_n > e^{a_n^2}+a_n > e^{a_n}$, contradiction. The red inequality is enough to prove that $\sum\frac{b_n}{a_n}$ is convergent, since it gives $\frac{b_n}{a_n}\leq a_n$ ($b_n$ cannot be negative since $e^x\geq x+1$).
So we just need to prove $(1)$. We can notice that over $(0,1]$ we have $x+e^{x^2} > 1+x+x^2$ and: $$1+x+x^2 > e^{x} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\ldots \tag{2}$$ since: $$1 > \sum_{n\geq 2}\frac{1}{n!} = e-2.\tag{3}$$
With the same argument we can also prove the stronger statement: $$\sum\frac{b_n}{a_n}\leq \frac{1}{e-1}\sum a_n.\tag{4}$$