Leibniz convergence test to compute the limit Can we compute the limit of a series using Leibniz test?
This is the problem I am struggling with:
" Let $(a_{n})_{n \ge 1}$ be a sequence of natural numbers, $a_{n} \ge 2$ ,
Let $b_{n} = 1 - \frac{1}{a_{1}} + \frac{1}{a_{1}a_{2}} +- \dots + (-1)^n\frac{1}{a_1a_2...a_n}$      , $n = 1,2,3, \dots$
Prove that:
a) $(b_n)_{n \ge 1}$ is convergent
b) if $(a_n)_{n \ge 1}$ is unbounded, then the limit of $(b_n)_{n \ge 1} \in \mathbb{R\setminus Q}$ "
(Source: Romanian National Olympiad Shortlist)
It's easy to prove point a) using Leibniz test. I may be wrong, but I belive the test can be used in point b), to prove the irrationality. Is there any way to compute the limit?

Here is my solution, but I am not sure if it is correct.
Suppose that $b_n \rightarrow \frac{p}{q}$ , where $p,q \in \mathbb{Z^*}$.
For any $\epsilon > 0$ , we have $|b_n - \frac{p}{q}| < \epsilon$ , and by choosing $\epsilon := \frac{\epsilon}{|q|}$ , we get:
$|1 - \frac{1}{a_1} + \frac{1}{a_1a_2} + \dots (-1)^n\frac{1}{a_1a_2...a_n} - \frac{p}{q}| < \frac{\epsilon}{|q|}$
Multiplying both sides with $|q|a_1a_2...a_n$ , we get:
$|q(a_1a_2...a_n - a_1a_2...a_{n-1} + \dots (-1)^n) - p| < \epsilon$ 
By setting $d_n = a_1a_2...a_n - a_1a_2...a_{n-1} + \dots (-1)^n$, we get that $d_n \rightarrow \frac{p}{q} \in \mathbb{Q}$
Since $a_n \ge 2$ for any $n \ge 1$ we get that $a_1a_2...a_n \ge 2^n$ , which implies $a_1a_2...a_n \rightarrow \infty$
Because $d_n = a_1a_2...a_{n-1}(a_n - 1) + a_1a_2...a_{n-3}(a_{n-2}-1) + \dots$ and $a_n - 1 \ge 1$ ,
we get that $d_n \rightarrow \infty$, contradiction
Can anybody tell me if this is correct?
