The unique positive real root of summation function

I post this question in MSE a week ago. I thought this should be an easy freshman exercise, but it turns out not easy...

The original question is very complicated, involving Bounded variation and other stuff. But I managed to simplify it into a simple algebra question, which stated below.

Define for $f(x)$, $x\geq 0$, (I don't care what happend for $x\leq 0$) $$f(x):=\sum_{k=1}^\infty \frac{a_k^2\lambda_k^2x}{(1+x\lambda_k)^2} - \sum_{k=1}^\infty \frac{b_k^2\beta_k}{(1+x\beta_k)^3}$$ where $a_k\in \mathbb R$, $b_k\in\mathbb R$, $\lambda_k>0$, $\beta_k> 0$, and $$\sum_{k=1}^\infty b_k^2< \sum_{k=1}^\infty a_k^2<\infty\,\text{ and }\sum_{k=1}^\infty a_k^2 \lambda_k< \sum_{k=1}^\infty b_k^2 \beta_k<\infty.$$ Additional assumption: we may think each $\beta_k$ is very large. You may take it as large as you want.

I am trying to prove that $f(x)$ has following graph. That is, prove that there exists $x_0>0$ such that $f(x_0)=0$, and for all $x<x_0$, $f(x)<0$, and for all $x>x_0$, $f(x)>0$.

Also, feel free to add whatever assumption on $a_k$, $b_k$, $\lambda_k$ and $\beta_k$ which you think won't trivial the question but reasonable. I wish to have at least some of them can work

Moreover, I may assume that $k$ is finite. i.e., we only have finite sum in $f$, not infinite.

• Your question is an analysis question, not algebra, nor number theory. I edited the tags accordingly. – GH from MO Feb 9 '16 at 1:26
• @GHfromMO Thank you for assigning the correct label! – JumpJump Feb 9 '16 at 2:16

I think, this is not true. We fix large $T$ and small $\varepsilon>0$ and denote $a(x)=x/(x+1)^2$, $b(x)=1/(x+T)^3$, $c(x)=1/(x+\varepsilon)^3$. We try to find for two positive numbers positive coefficients $A,B,C$ such that $$Aa(x)-Bb(x)-Cc(x)=Aa(y)-Bb(y)-Cc(y)=0$$ for two positive numbers $x>y$, say $x=10,y=9$. Considering this as a linear system in $A,B,C$ we find $A=b(x)c(y)-c(x)b(y)$, $B=a(x)c(y)-a(y)c(x)$, $C=a(y)b(x)-a(x)b(y)$. Thus we need $$\frac{b(x)}{b(y)}> \frac{a(x)}{a(y)}> \frac{c(x)}{c(y)}.$$ For large $T$ the ratio $b(x)/b(y)$ tends to 1, for small $\varepsilon$ the ratio $c(x)/c(y)$ tends to $y^3/x^3$. Both inequalities $a(y)>a(x)$ and $x^3a(x)>y^3b(y)$ are easy to check.
On the other hand, the statement is true if suppose that $\min \beta_i\geqslant \max \lambda_i$. Indeed, it is easy to see that the function $h_{\beta,\gamma}(t):=(1+\beta t)/(1+\gamma t)$ decreases on $(0,+\infty)$ for $0<\beta<\gamma$ and increases for $0<\gamma<\beta$. Hence if $\lambda_i\leqslant \beta\leqslant \beta_i$, the functions $x(1+\beta x)^3/(1+\lambda_i x)^2=x(1+\lambda_i x)(h_{\beta,\lambda_i}(x))^3$ increase and functions $(1+\beta x)^3/(1+\beta_i x)^3=(h_{\beta,\beta_i}(x))^3$ decrease. Totally, we see that $(1+\beta x)^3 f(x)$ is a sum of increasing summands, hence it increases, hence has unique root as desired (the existence of a root follows from $f(0)<0$, $\lim_{x\rightarrow +\infty} x^2 f(x)=+\infty$ and intermediate value theorem.)
• Sorry, I am a little bit confused. What is your $A$ $B$ $C$ and $a$ $b$ $c$ in my question? – JumpJump Feb 9 '16 at 13:34
• $\lambda_1=1$, $A=a_1^2$, $\beta_1=1/T$ and so on. – Fedor Petrov Feb 9 '16 at 13:51
• $B=(Tb_1)^2$, $\beta_2=1/\varepsilon$, $C=(Tb_2)^2$ – Fedor Petrov Feb 9 '16 at 14:15
• The $\beta_k$ in my problem is actually very large, so it may disprove your counterexample. (your counterexample is indeed very good, give me the idea what's going on here.) – JumpJump Feb 9 '16 at 15:04