Problematic Product Series!

Here's one solution. It's a tad long so we'll split it into three parts. We'll use an inductive argument to show that our product series $P$ is just another way of writing

$\ln(2) = \sum_{k = 1}^{k \rightarrow \infty} \frac{(-1)^{k+1} }{k}$ and then everything will fall into place nicely.

Part One : Rewriting the Product Series.

Consider a single factor of our product series $(1 + \frac{(-1)^{k+1}(k-1)!}{a_k} )$ . Re-arranging we get $(1 + \frac{(-1)^{k+1}(k-1)!}{a_k} ) = (\frac{a_k + (-1)^{k+1}(k-1)!}{a_k }) = \frac{a_{k+1}}{a_k (k + 1)}$ by recalling the definition of $a_{n+1}$ .

Now let $P(n) = \prod_{k = 2}^n ( 1 + \frac{(-1)^{k+1}(k-1)!}{a_k}).$ This is the partial product of $P$ . Using what we learnt about each factor we get,

$P(n) = \frac{a_{3}}{a_2 (3)} \times \frac{a_{4}}{a_3 (4)} \times ... \frac{a_{n}}{a_{n-1} (n )} \times \frac{a_{n+1}}{a_n (n + 1)}$ and by cancelling we get

$= \frac{a_{n+1}}{a_2 \times 3 ... \times (n+1)} = \frac{a_{n+1}}{(n+1)!}$ , remembering that $a_2 = 2$ by definition.

Part Two : Proof by Induction.

Let $T(n) = \sum_{k=1}^{k=n}\frac{(-1)^{k+1}}{k}.$ Now we'll show by induction on $n$ that $P(n) = T(n)$ for all natural numbers $n \ge 2.$

Base Case

When $n = 2$ we have $P(2) = 1 + \frac{(-1)^3 (1!)}{2} = \frac{1}{2} = \frac{1}{1} - \frac{1}{2} = T(2).$

Inductive Step

We'll assume that $T(k) = P(k)$ for some positive integer $k \ge 2$ and show that this implies that $T(k+1) = P(k+1).$

Using what we learnt in Part One , we have

$P(k+1) = P(k)(1 + \frac{(-1)^{k+2}k!}{a_{k+1}}) = \frac{a_{k+1}}{(k+1)!} (1 + \frac{(-1)^{k+2}k!}{a_{k+1}})$ $= \frac{a_{k+1}}{(k+1)!} + \frac{a_{k+1}(-1)^{k+2}k!}{a_{k+1}(k+1)!} = P(k) + \frac{(-1)^{k+2}}{k+1} = T(k) + \frac{(-1)^{k+2}}{k+1} = T(k+1)$ which is what we wanted.

So $T(n) = P(n)$ for all $n \ge 2$ .

Part 3 : Finishing up.

It follows that in this case $P = \lim_{n\rightarrow \infty} P(n) = \lim_{n\rightarrow \infty} T(n) = \sum_{n = 1}^{n \rightarrow \infty} \frac{(-1)^{n+1} }{n} = \ln(2)$ .

As a result the answer is $e^P = e^{\ln(2)} = 2.$