Convergence of $ \sum_{k=1}^{\infty} \sqrt{a} \prod_{i=1}^{k}\frac{1}{1+i a}$ as $a \rightarrow 0$

Question

Using numerical simulation, I can see that $$ v(a)=\sum_{k=1}^{\infty} \sqrt{a} \prod_{i=1}^{k}\frac{1}{1+i a} $$ converges to some value $1<v(a)<2$ as $a \rightarrow 0$. However, I couldn't find bounding series to prove it. Can anybody help? Thanks!

Answer 1

I will denote $z = a^{-1}$ and $f(z) = v(a)$. Then we have

\begin{align*} f(z) &= z^{-1/2} \sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{z}{k + z} = z^{-1/2} \sum_{n=1}^{\infty} z^{n}\frac{\Gamma(z+1)}{\Gamma(z+n+1)} \\ &= z^{-1/2} \sum_{n=1}^{\infty} \frac{z^{n}}{\Gamma(n)} \int_{0}^{1} x^{z}(1-x)^{n-1} \, dx \\ &= z^{1/2} \int_{0}^{1} x^{z} e^{z(1-x)} \, dx \\ &= \frac{e^{z}}{z^{z+1/2}} \int_{0}^{z} t^{z} e^{-t} \, dt. \qquad (t = zx) \end{align*}

Now by noting that $ z! \sim \sqrt{2\pi} z^{z+1/2} e^{-z}$ and

$$ \int_{0}^{z} t^{z} e^{-t} \, dt \sim \frac{z!}{2} $$

as $z \to \infty$, it follows that

$$ \lim_{a\to 0^{+}} v(a) = \lim_{z\to\infty} f(z) = \sqrt{\frac{\pi}{2}} \approx 1.2533141373155002512\cdots. $$