Let $\lambda>0$ be fixed, and $a,b>0$ positive real numbers. We have a series which is defined as
$$\sum_{n=0}^{\infty}\bigg(\prod_{j=1}^{n}\frac{1}{1+\lambda\frac{1}{a+b(j-1)}}\bigg).$$
Is there any way to choose $a$ and $b$ to make this series convergent? If this series diverges for all $a, b>0$, how should we prove it?
Many thanks for any hints and solutions!
Hint: it suffices to show that there exists a constant $\alpha>1$ such that $\displaystyle\sum_{j=1}^n \log\bigg(1+\lambda\frac1{a+b(j-1)}\bigg) > \alpha\log n$ for sufficiently large $n$. (exponentiate, take reciprocals ...) But $\displaystyle\log\bigg(1+\lambda\frac1{a+b(j-1)}\bigg) \approx \lambda\frac1{a+b(j-1)}$ when $j$ is large; and comparing $\displaystyle\sum_{j=1}^n \lambda\frac1{a+b(j-1)}$ to an integral shows that $\displaystyle\sum_{j=1}^n \lambda\frac1{a+b(j-1)} \sim \frac\lambda b\log n$. So it suffices (and indeed is perhaps necessary as well) to take $\displaystyle b<\frac1\lambda$.