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QuestionFind the sum of the series ∑n=1∞1n(n+2)\sum_{n=1}^{\infty} \frac{1}{n(n+2)}.

Studdy Solution

STEP 1

Assumptions1. We are asked to evaluate the infinite series. . The series is given by ∑n=1∞1n(n+)\sum_{n=1}^{\infty} \frac{1}{n(n+)}

STEP 2

This is a telescoping series, which means that we can rewrite each term such that it is the difference between two successive terms of a sequence.
The general term of the given series can be rewritten as below1n(n+2)=An+Bn+2\frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}

STEP 3

To find A and B, we must solve the following system of equations{A+B=0 (by comparing coefficients of n0)2A=1 (by comparing coefficients of n−1)\begin{cases} A+B =0\ (by\ comparing\ coefficients\ of\ n^0) \\ 2A =1\ (by\ comparing\ coefficients\ of\ n^-1) \end{cases}

STEP 4

olving the system of equation, we have{A=12B=−12\begin{cases} A=\frac{1}{2} \\ B=\frac{-1}{2} \end{cases}

STEP 5

Substitute A and B back into the expression for each term in the series1n(n+2)=1/2n−1/2n+2\frac{1}{n(n+2)} = \frac{1/2}{n} - \frac{1/2}{n+2}

STEP 6

Now, for the sum ∑n=1∞1n(n+2)\sum_{n=1}^{\infty} \frac{1}{n(n+2)}, we write out the first few terms and observe the pattern1/21−1/23+1/22−1/24+1/23−1/25+…\frac{1/2}{1} - \frac{1/2}{3} + \frac{1/2}{2} - \frac{1/2}{4} + \frac{1/2}{3} - \frac{1/2}{5} + \ldots

STEP 7

We obtain a telescoping sum, where each term cancels out the previous one (after the first), and as n approaches infinity, the last terms also get cancelled.
So the final sum is1/21+1/22=12+14=34\frac{1/2}{1} + \frac{1/2}{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} Therefore, the sum of the series ∑n=1∞1n(n+2)\sum_{n=1}^{\infty} \frac{1}{n(n+2)} is 34\frac{3}{4}.

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