1
$\begingroup$

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?

  • 1
    @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.)2011-10-29

3 Answers 3

1

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 $a,b >0$

7

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\}}. $

  • 0
    @robjohn, thanks. As the Edit shows, even this is not needed.2011-10-29
4

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 $(2)$.