Can you prove the answer?

Note that $(a_n - 1)(b - 1 + a_{n-1}) + 1 \; = \; 0$ If $\sum_n a_n$ converges, then $\lim_{n\to\infty}a_n = 0$ . The above identity then shows that $-(b-1) + 1 \; = \; 0$ and hence $b=2$ . But then $0 \; = \; (a_n-1)(a_{n-1}+1)+1 \; = \; a_na_{n-1} - a_{n-1} + a_n$ and hence $a_n^{-1} \; = \; a_{n-1}^{-1} + 1$ so that $a_n^{-1} = n$ , and so $a_n = n^{-1}$ . But then $\sum_n a_n$ diverges.

Thus there is no value of $b$ (integer or otherwise) for which $\sum_n a_n$ converges. It is worth noting that there are some values of $b$ for which the sequence is not defined. When $b=1$ , for example, we have $a_2=0$ , and so $a_3$ does not exist.