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Let $$\begin{align*} \sum u_{n},\sum v_{n} \end{align*}$$ be two positive convergent series. How can I prove that $$\begin{align*} \sum \frac{u_{n}v_{n}}{au_n+bv_n}; a,b>0 \end{align*}$$ converge?

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Hint: Multiplication goes faster to zero than addition. – Asaf Karagila Oct 29 '11 at 11:07
Take any integer $n$, then in case $0< u_n\leq v_n$ you'll have (using positivity of $u,v,a$ and $b$) $$\frac{u_nv_n}{au_n+bv_n}\leq\frac{u_nv_n}{au_n+bu_n}=\frac{v_n}{a+b}$$ and in case $0< v_n\leq u_n$ you similarly have $$\frac{u_nv_n}{au_n+bv_n}\leq\frac{u_n}{a+b}.$$ Either way you have $$0<\frac{u_nv_n}{au_n+bv_n}\leq\frac{u_n+v_n}{a+b}$$ where the right hand side defines a convergent series. – Olivier Bégassat Oct 29 '11 at 11:11
31 minutes. $ $ – Did Oct 29 '11 at 11:36
@DidierPiau I don't understand what's going on ^^ what "turned around" at 31 minutes? – Olivier Bégassat Oct 29 '11 at 12:10
@Olivier, an answer got accepted 31 minutes after the question was posted. This behaviour seems to be more and more common amongst certain MSE posters although it has some obvious drawbacks, to which I wish to draw attention. For example, some would-be answerers will be de facto unable to post answers until after one got accepted, if only for time zone reasons. (The present post is a good case to mention this because the accepted answer is pretty decent.) – Did Oct 29 '11 at 13:10
up vote 4 down vote accepted

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 $(2)$.

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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 $n$, $$ \frac{\min\{u_n,v_n\}}{a+b}\leqslant w_n\leqslant\frac{\min\{u_n,v_n\}}{\min\{a,b\}}. $$

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There is something elegant about the convergence of only one sequence being needed. (+1) – robjohn Oct 29 '11 at 11:50
@robjohn, thanks. As the Edit shows, even this is not needed. – Did Oct 29 '11 at 12:01

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 $a,b >0$

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