If $S_n=\sum _{i=1}^n\frac{1}{a_i}$ and $a_k=\left(\prod _{i=1}^{k-1}a_i\right)+1$ then evaluate $\lim _{n\to \infty }S_n$.

Question

If $$S_n=\sum _{i=1}^n\frac{1}{a_i}$$ and $$a_k=\left(\prod _{i=1}^{k-1}a_i\right)+1$$ then evaluate $$\lim _{n\to \infty }S_n$$ I have tried to simplify a few terms, but it does not seem to cancel out nicely.

Also by evaluating a few terms, it never appears to cross 1 and always gets closer towards it.

A hint on how to solve this problem will be greatly appreciated.

Thanks☺

Answer 1

Outline:

• Show, by induction on $n$, that $$ S_n = \frac2{a_1} - \frac1{a_{n+1}-1} $$ for all $n\ge1$. (This is straightforward once we note that $a_n(a_n-1) = a_{n+1}-1$.)
• Show that $\lim_{n\to\infty} a_n = \infty$. (This requires the unstated assumption that $a_1>0$, and is less trivial than it looks when $a_n<1$, but it's still true.)

These two steps combine immediately to give $\lim_{n\to\infty} S_n = 2/a_1$.

(Side note: when $a_1=2$ this is the famous Sylvester's sequence, so this problem is a generalization of the fact that Sylvester's sequence was constructed so that the sum of the reciprocals of the terms in the sequence is equal to $1$.)