The context is analysing the behaviour of the Prod algorithm under a certain input. I have two products
$$W_1^n = \prod_{i=1}^n\left(1 + \frac{\alpha(-1)^i}{ \sqrt i}\right) \qquad \qquad \qquad W_2^n = \prod_{i=1}^n\left(1 + \frac{\alpha(-1)^i}{ i}\right)$$
for some $\alpha < 1$. These are used to construct the point $\frac{(W_1,W_2)}{W_1+W_2}$ in the simplex.
I don't know much about analysing products in the first place. Are there any obvious tips or tricks that look useful for the following problem?
If I plot the products it looks like both converge to some $W_1^n \to W_1$ and $W_2^n \to W_2$ as $n \to \infty$. Moreover the sign changes ensures that $W_1^n$ is alternately greater and less than the limit. I am interested in bounding from below the size $|W_1^n - W_1|$ and $|W_2^n - W_2|$ of these oscillations. Experimentally it seems they have size $O(1/\sqrt n)$ and I would like to prove that.
It is believable we could estimate $W_1^n$ as follows: $(1)$ Expand log $W_1^n$ as a sum. $(2)$ Separate the positive and negative parts of the sum. $(3)$ Use convexity to bound the positive and negative parts from both sizes. $(4)$ Replace the resulting sums over $1/\sqrt n$ or $1/n$ with integrals. $(5)$ use the integrals to estimate the sums from both sides.
However this sounds like an awful lot of work and might not be helpful if the errors in the estimate are larger than the size of the oscillations.
$$\left(1-\frac{\alpha}{2k-1}\right)\left(1+\frac{\alpha}{2k}\right)=1-\frac{\alpha(1+\alpha)}{2k(2k-1)}$$ easily implies that $W_2$ converges (and the bounds are easier to obtain just this way, i.e. not by separating the parts, but by estimating the product of even (or odd) number of terms, and adding the extra term if any).
But for $$a_k:=\left(1-\frac{\alpha}{\sqrt{2k-1}}\right)\left(1+\frac{\alpha}{\sqrt{2k}}\right)=1-\frac{\alpha(\alpha+\sqrt{2k}-\sqrt{2k-1})}{\sqrt{2k(2k-1)}}$$ we have $k(1-a_k)\underset{k\to\infty}{\longrightarrow}\alpha^2/2$, thus $W_1$ diverges (to zero) if $\alpha\neq 0$.
A side note. $W_2$ can be computed using the gamma function: $$\prod_{k=1}^{\infty}\left(1+\frac{\alpha(-1)^k}{k}\right)=\frac{\left(\prod_{k=1}^{\infty}(1+\frac{\alpha}{2k})\exp(-\frac{\alpha}{2k})\right)^2}{\prod_{k=1}^{\infty}(1+\frac{\alpha}{k})\exp(-\frac{\alpha}{k})}=\frac{\Gamma(1+\alpha)}{\Gamma^2(1+\alpha/2)}=\ldots$$