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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

S_n=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{n(n+1)(n+2)}

, then

8S_\infty

2

correct

4

6

8

Solution

Step 1: Split each term:

\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:

S_n=\frac12\left[\frac{1}{1\cdot2}-\frac{1}{(n+1)(n+2)}\right].

Step 3: As

n\to\infty

the second term vanishes:

S_\infty=\frac12\cdot\frac12=\frac14\ \Rightarrow\ 8S_\infty=2.

Correct answer: (1)

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