Solving a recurrence: $\frac{1}{a_{n}}=\frac{1}{a_{n-1}}+\frac{1}{a_{n+1}}$

Question

As in the title, solve such recurrence: $$\frac{1}{a_{n}}=\frac{1}{a_{n-1}}+\frac{1}{a_{n+1}}$$ for $n\ge 2$, where $a_1=2$ and $a_2=1$.

I mean, any hints?

Answer 1

Hint: Can you solve $$b_{n+1} = b_n - b_{n-1}\qquad n\geq 2$$ where $b_1=\frac{1}{2}$, $b_2=1$?

If so, can you define $(b_n)_n$ with regard to $(a_n)_n$ so that solving the above is equivalent to solving your original problem?