Student Number:_{} No._{} Marks _{}

Solution. Let a_{n} = [1/((n+1)2^{n})] and a_{n+1} = [1/((n+2)2^{n+1})]. Then [(a_{n})/(a_{n+1})] = [(2(n+1))/(n+1)] and

Hence, we have
(i) when x1 < 2 (or 1 < x < 3), the series converges absolutely.
(ii) When x1 > 2 (or x < 1 or x > 3), the series diverges.
(iii) When x1 = 2 (or x = 3), the series becomes å_{n = 0}^{¥}[1/(n+1)] and diverges.
(iv) When x1 = 2 (or x = 1), the series becomes å_{n = 0}^{¥}[((1)^{n})/(n+1)]. It follows from the Alternating Series Test that å_{n = 0}^{¥}[((1)^{n})/(n+1)] converges.
Hence, the convergence set is 1 £ x < 3 or [1, 3).