ChallengeChallengeTelescoping & Recursion ShareTelescoping & Recursion · Challenge problemA sequence is defined by a1=53a_1 = \dfrac{5}{3}a1=35 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?A52\dfrac{5}{2}25B12\dfrac{1}{2}21C32\dfrac{3}{2}23D333E−32- \dfrac{3}{2}−23Check Show answer Walk through it step by step Practice more like this Review this concept