# Estimate a sum of products

Let $0 < \alpha < 1$. Show that for $\lambda > 0$ big enough $$\sum_{n=1}^\infty \prod_{k=1}^n \frac{1}{\lambda k^{-\alpha} + 1} < \infty$$

I think $\lambda = 1$ is enough. You could then use the estimation \begin{align*} \prod_{k=1}^n \frac{1}{ k^{-\alpha} + 1} &\le \prod_{k=1}^n \frac{1}{ n^{-\alpha} + 1} \\ &= \left(\frac{1}{ n^{-\alpha} + 1}\right)^n \\ &= \left(1-\frac{1}{ 1+ n^{\alpha}}\right)^n \\ \end{align*} which looks a bit like the product respresentation of $\exp$.

I've checked it numerically. For small $\alpha$ the expression seems to get quite big, so I'm not 100% sure.

-
@Haderlump: Sorry, I misread (should not have deleted my comment, so you could see!) I thought it was a $+$. We have $\left(1-\frac{1}{n^\alpha+1}\right)^n$. The power can be written as $(n^\alpha+1)(n/(n^\alpha+1))$. So after a while we are looking at $d^{n/(n^\alpha+1)}$ where $d$ is close to $1/e$, and in particular $\lt 1/2$. –  André Nicolas Oct 7 '12 at 17:53
We have the inequality $\log(1-x)\leq -x$ for $0\leq x\leq 1$. Indeed, defining $f(x):=\log(1-x)+x$ we get $f'(x)=1-\frac 1{1-x}=-\frac x{1-x}\leq 0$ so $f(x)\leq f(0)=0$. This gives $$n\log\left(1-\frac 1{1+n^{\alpha}}\right)\leq -\frac n{1+n^{\alpha}}$$ and $$\exp\left(n\log\left(1-\frac 1{1+n^{\alpha}}\right)\right)\leq \exp\left( -\frac n{1+n^{\alpha}}\right)=\exp\left(-n^{1-\alpha}\frac{n^{\alpha}}{1+n^{\alpha}}\right).$$ We choose $N$ such that for $n\geq N$, $\frac{n^{\alpha}}{1+n^{\alpha}}\geq\frac 12$. Then for such $n$, $$\left(1-\frac 1{1+n^{\alpha}}\right)^n\leq \exp(-n^{1-\alpha}),$$ and the series $\sum_{n\geq 1}\exp(-n^{1-\alpha})$ is convergent.