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I need to compute the sum of the following infinite series:

$$\frac{1}{1 \cdot 3} + \frac{1}{2\cdot 4} + \frac{1}{3\cdot 5} + \frac{1}{4\cdot 6} +\cdots $$

How would I compute this sum? My teacher didn't really go over telescopic sums, and in the book it says I need to find the closed form of this sum, which I can do and I did.

$1/(n)(n+2)$

From here on though I get really confused, the book separates the fraction somehow and then tries to list the terms in closed form and factor something out? Not sure.

Zion Todd
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2 Answers2

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Note that the general term is

$$\frac{1}{n(n+2)} = \frac{1}{2}\left( \frac 1n - \frac{1}{n+2}\right).$$

Try to write down the first several terms and see how things cancel each other.

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$$\sum_{n=1}^k \frac{1}{n(n+2)}=\frac{1}{2}\sum_{n=1}^k \left(\frac{1}{n}-\frac{1}{n+2}\right)$$

$$=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2}\right)$$

Your answer will be $\displaystyle{\lim_{k\to +\infty}\left(\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2}\right)\right)}$.

user236182
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