Question

Find the sum of series 1/2 + 1/6 + 1/12 + ------1/n(n+1) terms

Answer 1

Answer:

n/(n + 1)

Step-by-step explanation:

General term = 1/n(n + 1)

    = ((n + 1) - n)/n(n + 1)

    = 1/n - 1/(n + 1)

Therefore, let the sum of the series be R.

⇒ 1/2 + 1/6 + 1/12 + ... 1/m(m + 1) = R

⇒ (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... [1/(n - 1) - 1/n] + [1/n - 1/(n + 1)] = R

⇒ 1 - 1/(n + 1) = R

⇒ (n + 1 - 1)/(n + 1) = R

⇒ n/(n + 1) = R  

   Hence, the sum of 1st n terms of this series is n/(n + 1).

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Answer 2

Answer:

What the OP is asking about is this sum:

∑n=1M1n(n+1)=∑n=1M1n−1n+1∑n=1M1n(n+1)=∑n=1M1n−1n+1

Specifically, the OP has M=49M=49, but that’s not important. What is important is that this is a telescoping series, so it simplifies to:

∑n=1M1n(n+1)=11−12+12−13+…+1M−1M+1=1−1M+1∑n=1M1n(n+1)=11−12+12−13+…+1M−1M+1=1−1M+1

Finally, plugging in M=49M=49, we get:

∑n=1491n(n+1)=1−150=4950=0.98∑n=1491n(n+1)=1−150=4950=0.98

And if we let M→∞M→∞, then the series limits to one.

Hope it will help you..