If the positive series $\sum u_n$ and $\sum v_n$ both converge then $\sum \frac{u_nv_n}{au_n+bv_n}$ converges, for every positive $a$ and $b$ Assume that $\sum u_n$ and $\sum v_n$ are two positive convergent series.
How to prove that, for every $a>0$ and $b>0$, the series $$\sum \frac{u_nv_n}{au_n+bv_n}$$
converges?
 A: Let $w_n=\dfrac{u_nv_n}{au_n+bv_n}$. Since $au_n+bv_n\geqslant au_n$, $w_n\leqslant\dfrac{v_n}a$. Likewise,  $au_n+bv_n\geqslant au_n$ hence $w_n\leqslant\dfrac{u_n}b$. 
Hence the series $\sum\limits_n w_n$ converges as soon as at least one of the two series $\sum\limits_n u_n$ and $\sum\limits_n v_n$ converges.

Edit By the argument above, the series $\sum\limits_n w_n$ may converge even when the series $\sum\limits_n u_n$ and $\sum\limits_n v_n$ both diverge. To see this, choose any convergent series $\sum\limits_n x_n$ with positive terms and define $(u_n)_n$ by $u_{2n}=1$ and $u_{2n+1}=x_{2n+1}$ and $(v_n)_n$ by $v_{2n}=x_{2n}$ and $v_{2n+1}=1$ for every nonnegative $n$. Then the series $\sum\limits_n u_n$ and $\sum\limits_n v_n$ both diverge because the series $\sum\limits_n1$ does while the series $\sum\limits_n w_n$ converges because the series $\sum\limits_nx_n$ does. 
This example may seem peculiar, in fact it captures the essence of what can happen. To wit, here is the last word on this problem:

The series $\sum\limits_n w_n$ converges if and only if the series $\sum\limits_n\min\{u_n,v_n\}$ does.

The proof is exceedingly simple: note that, for every $n$,
$$
\frac{\min\{u_n,v_n\}}{a+b}\leqslant w_n\leqslant\frac{\min\{u_n,v_n\}}{\min\{a,b\}}.
$$
A: Since $0\le au_n^2+bv_n^2$, add $(a+b)u_nv_n$ to both sides:
$$
(a+b)u_nv_n\le(a+b)u_nv_n+au_n^2+bv_n^2=(au_n+bv_n)(u_n+v_n)\tag{1}
$$
Divide both sides of $(1)$ by $(a+b)(au_n+bv_n)$ and you get
$$
\frac{u_nv_n}{au_n+bv_n}\le\frac{u_n+v_n}{a+b}\tag{2}
$$
Since both $\sum u_n$ and $\sum v_n$ are absolutely convergent, $\sum\frac{u_n+v_n}{a+b}$ is also. Therefore, $\sum\frac{u_nv_n}{au_n+bv_n}$ is convergent by the comparison test and $(2)$.
A: 1/ series $\ \sum max(u_{n}, v_{n})$ is convergent 
2/ $0< \frac{xy}{ax+by} < \frac{1}{2ab}max(x,y)$ for $x,y \geq 0 $ and $a,b >0$
