# If $\sum\frac{a_n}{b_n}$ and $\sum\frac{a_n^2}{b_n^2}$ both converge then $\sum\frac{a_n}{a_n + b_n}$ converges

If $\sum\limits_{n=0}^\infty\frac{a_n}{b_n}$ converges and $\sum\limits_{n=0}^\infty\frac{a_n^2}{b_n^2}$ converges, where $(a_n + b_n)b_n \ne 0$ for every $n \in \mathbb{N}$ , then show that $\sum\limits_{n=0}^\infty \frac{a_n}{a_n + b_n}$ also converges.

I have proved it for non-negative $a_n$ and $b_n$, but I am unable to do it for the other cases.

• You should learn how to write in LaTeX! – Mercy King Sep 1 '12 at 13:05
• Why do you need the condition $(a_n+b_n)b_n \ne 0$ for? – Mercy King Sep 1 '12 at 13:16
• To have all the sequences well-defined . – Ester Sep 1 '12 at 13:18
• Isn't $a_n+b_n \ne 0$ enough?! – Mercy King Sep 1 '12 at 13:28
• What about the other 2 sequences? – Ester Sep 1 '12 at 13:31

Considering $c_n=\frac{a_n}{b_n}$, one assumes that $C=\sum\limits_nc_n$ and $C'=\sum\limits_nc_n^2$ both converge and that $c_n\ne-1$ for every $n$, and one wants to show that $D=\sum\limits_nd_n$ converges, with $d_n=\frac{c_n}{1+c_n}$.
• Since $d_n=c_n-e_n$ with $e_n=\frac{c_n^2}{1+c_n}$ and $C$ converges, $D$ converges if and only if $E=\sum\limits_ne_n$ does.
• Since $C$ converges, $c_n\to0$ hence $|c_n|\leqslant\frac12$ for every $n$ large enough, and then $|1+c_n|\geqslant\frac12$.
• For every $n$ large enough, $\left|e_n\right|\leqslant2c_n^2$.
• Hence the convergence of $C'$ implies the (absolute) convergence of $E$. Done.