Limit of products in $x_n = 1-An^{-\alpha}$ and their summation

Question

Suppose that we have $A >0, \alpha >0$, and for each $n$, define $x_n = 1-An^{-\alpha}$ such that for large $n$ we have $x_n \in (0,1)$. Also, define the product sequence, $y_n = \prod_{i=0}^n x_i$. Show that

1. $\lim_{n\to\infty} y_n > 0$ if $\alpha > 1$.
2. Let $\alpha = 1$. Then, $\lim_{n\to\infty} \sum_{i=0}^n y_i < \infty$ if $A > 1$ and $ = \infty $ if $A \leq 1$.

I have tried to use the bound $1+x \leq e^x$ to approximate the upper bound for each term by far, but it does not seem to work. Finding exponential lower bounds for each term would make it easier to solve the first problem; however, now I wonder if this is a right way to do it.

Answer 1

Note: we want to start summation from $n=1$ or $n=2$, $2$ depending on $A$ (example: if $A=1$ then $x_1= 1-A 1^{-\alpha}=0$).

1. $(y_n)$ is decreasing.

Now

$$\ln y_n = \sum_{j\ne n} \ln (1-A n^{-\alpha}).$$

Use the inequality $\ln (1-x) > -2x$ for $x$ small enough, gives

$$ \ln y_n \ge -c \sum n^{-\alpha},$$

for some $c>0$. The righthand side converges and so $y_n$ is bounded below by a (strictly) positive constant.

2. Now $\alpha =1$.

On the one hand since $1-x \le e^{-x}$, we have that

$$y_n \le e^{ -A \sum_{j=1}^n 1/j}.$$

Since $\sum_{j=1}^n 1/j\sim \ln n$ as $n\to\infty$, for any given $\epsilon>0$, we have that for $n$ large enough the righthand side is bounded above by

$$ e^{-A(1-\epsilon) \ln n}= n^{-A(1-\epsilon)}.$$

Therefore if $A>1$, we can choose $\epsilon$ so that $A(1-\epsilon)>1$, and the series $\sum_{j=1}^\infty y_j $ converges by comparison with $\sum n^{-A(1-\epsilon)}$.

Next we will show that $\sum y_j$ diverges if $A=1$, and by monotonicity, this implies that the series diverges whenever $A<1$. We have

$$x_i = 1-\frac{1}{i} = (i-1)/i.$$

Therefore $(1-1/2)(1-1/3)\dots (1-1/j) =\frac{(j-1)!}{j!}=1/j$.

It follows that $\sum y_j$ is simply the harmonic series (excluding the first term).