Weird Recurrence, Weirder Sum

If we write $A(n) \; = \; \prod_{m=1}^n a_m \qquad \qquad n \ge 1$ then the recurrence relation can be rewritten as $\frac{1}{A(n-1)} \; = \; \frac{1}{A(n)} + \frac{1}{a_n} \qquad \qquad n \ge 2 \;,$ and so $\begin{array}{rcl} \displaystyle \sum_{n=1}^N \frac{1}{a_n} & = & \displaystyle \frac{1}{a_1} + \sum_{n=2}^N \left(\frac{1}{A(n-1)} - \frac{1}{A(n)}\right) \\ & = & \displaystyle \frac{1}{a_1} + \frac{1}{A(1)} - \frac{1}{A(N)} \; = \; \frac{2}{a_1} - \frac{1}{a_1a_2\cdots a_N} \end{array}$ which tends to $\frac{2}{a_1} = \boxed{1}$ as $N \to \infty$ .

From the first few $a_n$ , we can see that $S_n = \displaystyle \sum_{i=1}^n \frac 1{a_i} = \frac {a_n(a_n-1)-1}{a_n(a_n-1)}$ . Let us proof that the claim is true for all $n \ge 1$ by induction.

Step 1:

For $n=1$ , $S_1 = \dfrac {a_1(a_1-1)-1}{a_1(a_1-1)} = \dfrac {2(2-1)-1}{2(2-1)} = \dfrac 12 = \dfrac 1{a_1}$ as given. Therefore, the claim is true for $n=1$ .

Now, from $\displaystyle a_n = 1 + \prod_{i=1}^{n-1} a_i \implies a_{n+1} = 1 + \prod_{i=1}^n a_i = 1 + a_n \prod_{i=1}^{n-1} a_i = 1 + a_n (a_n -1)$

Step 2:

Assuming the claim is true for $n$ , then we have:

$\begin{aligned} S_{n+1} & = S_n + \frac 1{a_{n+1}} \\ & = \frac {a_n(a_n-1)-1}{a_n(a_n-1)} + \frac 1{a_{n+1}} \\ & = \frac {a_{n+1}-2}{a_{n+1}-1} + \frac 1{a_{n+1}} \\ & = \frac {a_{n+1}^2 - 2a_{n+1} + a_{n+1} -1 }{a_{n+1}(a_{n+1}-1)} \\ & = \frac {a_{n+1}(a_{n+1} -1) -1}{a_{n+1}(a_{n+1}-1)} \end{aligned}$

Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 1$ .

Now, we have:

$\begin{aligned} \lim_{n \to \infty} S_n & = \lim_{n \to \infty} \frac {a_n(a_n-1)-1}{a_n(a_n-1)} \\ & = \lim_{n \to \infty} \left(1 - \frac 1{a_n(a_n-1)} \right) & \small \color{#3D99F6}{\text{As }n \to \infty, \ a_n \to \infty} \\ & = \boxed{1} \end{aligned}$

This is one of my favorite sequences......It is known as Sylvester's Sequence