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Sum of 1/[n(n+1)(n+2)] to Infinity: 8 S-infinity | JEE

JEE Maths question with a full step-by-step solution.

Question
If Sn=11⋅2⋅3+12⋅3⋅4+⋯+1n(n+1)(n+2)S_n=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{n(n+1)(n+2)}, then 8S∞8S_\infty is
A22correct
B44
C66
D88
Solution
Step 1: Split each term:
1n(n+1)(n+2)=12[1n(n+1)−1(n+1)(n+2)].\frac{1}{n(n+1)(n+2)}=\frac12\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right].
 Step 2: Sum telescopes:
Sn=12[11⋅2−1(n+1)(n+2)].S_n=\frac12\left[\frac{1}{1\cdot2}-\frac{1}{(n+1)(n+2)}\right].
 Step 3: As n→∞n\to\infty the second term vanishes:
S∞=12⋅12=14 ⇒ 8S∞=2.S_\infty=\frac12\cdot\frac12=\frac14\ \Rightarrow\ 8S_\infty=2.
 Correct answer: (1)
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