I found this thread, but since it says that I can't ask for help there, I'm making a new one. How to find the partial sum of a given series?

I have this series (WolframAlpha): $$ \sum_{n = 3}^\infty \frac{1}{n(n - 2)} $$

And I need the partial sum formula. I don't get how it is derived. I got to the partial fraction decomposition described in the linked thread. I made it to the point where I ahd to put the newly found coefficients into the sum. This is what I got: (WolframAlpha). Ignore the output. Now how do I get from it to the partial sum formula in the first Wolfram Alpha link?


1 Answer 1


Set $$a_n := \frac{1}{n(n - 2)} = \frac{1}{2}(\frac{1}{n - 2} - \frac{1}{n})$$ and $S_m := \sum_{n = 3}^m a_n$ for $n, m \geq 3$. From the partial fraction decomposition, the middle parts cancel each other. For example, $$ \begin{align*} S_5 &= \frac{1}{2}\Big((\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5})\Big) \\ &= \frac{1}{2} \Big( \frac{1}{1} + \frac{1}{2} - \frac{1}{4} - \frac{1}{5}\Big). \end{align*}$$

Similarly, we get $$ \begin{align*} S_m &= \frac{1}{2} \Big(\frac{1}{1} + \frac{1}{2} - \frac{1}{m - 1} - \frac{1}{m} \Big) \\ &= \frac{3m^2 - 7m + 2}{4m(m - 1)}. \end{align*}$$


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