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How do I find a formula for the partial sum $s_n$ for the series $$\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+2})$$

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closed as off-topic by Did, Your Ad Here, egreg, MathOverview, Davide Giraudo Mar 25 '14 at 17:03

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Did you try to compute $s_n$ for some values of $n$, to see what happens? – Did Mar 25 '14 at 15:14
You sum turns out to be$$\dfrac{n(3n+5)}{2(n+1)(n+2)}$$ – Sanath K. Devalapurkar Mar 25 '14 at 15:17
@Ale Yep!${}{}$ – Sanath K. Devalapurkar Mar 25 '14 at 15:25
up vote 6 down vote accepted

Hint Limit the sum by some $N$ and separate it, changing the index: $$ \sum_{k=1}^N \left( \frac{1}{k} - \frac{1}{k+2} \right) = \sum_{k=1}^N \frac{1}{k} - \sum_{k=1}^N \frac{1}{k+2} = \sum_{k=1}^N \frac{1}{k} - \sum_{k=3}^{N+2} \frac{1}{k} $$ can you take it from here?

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For example: $$\left(\dfrac{1}{1} - \dfrac{1}{3}\right) + \left(\dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{3} - \dfrac{1}{5}\right) = \dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{4} - \dfrac{1}{5}$$ Try a few more: do you see the pattern?

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Hint: Try to figure out the cancellations happening in

$$f(1) {\color{red}{- f(3)}} + f(2) \color{blue}{- f(4)} \color{red}{+ f(3)} \color{blue}{- f(4)} +\cdots\\+f(n-2) - f(n) + f(n-1) - f(n+1) + f(n) - f(n+2)$$

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