ChallengeChallengeTelescoping & Recursion ShareTelescoping & Recursion · Challenge problemA sequence is defined by a1=32a_1 = \dfrac{3}{2}a1=23 and an+1=an2−an+1a_{n+1} = a_n^2 - a_n + 1an+1=an2−an+1 for n≥1n \ge 1n≥1. What is ∑n=1∞1an\displaystyle\sum_{n=1}^{\infty} \frac{1}{a_n}n=1∑∞an1?A333B111C444D222E−2-2−2Check Show answer Walk through it step by step Practice more like this Review this concept