How to evaluate $\sum_{n=1}^{\infty} \frac{1}{a_n}$

Question

If $a_1=2$ and $a_{n+1}=a_n^2-a_n+1$ for $n\geqslant1$ then how do I find $$\sum_{n=1}^{\infty} \frac{1}{a_n}.$$ Any hint is enough. Thanks.

Answer 1

We have that $a_{n+1}-1=a_n^2-a_n$ and $$\frac{1}{a_{n+1}-1}=\frac{1}{a_n^2-a_n}=\frac{1}{a_n(a_n-1)}=\frac{1}{a_n-1}-\frac{1}{a_n}\implies \frac{1}{a_n}=\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}.$$ Hence, the sum is telescoping: $$\sum_{n=1}^{N} \frac{1}{a_n}=\sum_{n=1}^{N} \frac{1}{a_n-1}-\sum_{n=1}^{N}\frac{1}{a_{n+1}-1}=\sum_{n=1}^{N} \frac{1}{a_n-1}-\sum_{n=2}^{N+1}\frac{1}{a_{n}-1}=\frac{1}{a_1-1}-\frac{1}{a_{N+1}-1}.$$ Can you take it from here?

Answer 2

Hint:

$$\sum_{k=1}^{n} \frac{1}{a_k} = 1-\frac{1}{a_{n+1}-1}.$$

See if you can prove this by induction.

Note: I found this identity by listing the first few partial sums.