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I'm quite stuck with the following problem. I have seen on this forum that there is already an answer for the infinite sum to the problem but I can't seem to find how to find the sum for a finite value.

The first part of the questions asks to transform the given series using partial fractions, which I did as follows:

$$ \frac{1}{k(k + 2)} $$

Which becomes:

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

The question now asks to evaluate the finite sum:

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

I have tried expanding the summation and I have been able to cancel out some terms, but I cannot seem to find a correct solution in the end. Has anyone any idea or method on how to evaluate these sums after rewriting them using partial fractions?

Thanks in advance!

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For instance, with $n = 10$ we have $$ \sum_{k=1}^n \frac1k - \frac 1{k+2} = \\ \left(1 - \frac 13 \right) + \left(\frac12 - \frac 14 \right) + \left(\frac13 - \frac 15 \right) + \cdots + \left(\frac 18 - \frac 1{10} \right) + \left(\frac19 - \frac 1{11} \right) + \left(\frac1{10} - \frac 1{12} \right) =\\ 1 + \frac 12 + \left(\frac13 - \frac 1{3} \right) + \cdots + \left(\frac1{10} - \frac 1{10} \right) - \frac 1{11} - \frac 1{12}. $$ For a more formal approach, note that $$ \sum_{k=1}^n \frac1k - \frac 1{k+2} = \sum_{k=1}^n \frac1k - \sum_{k=1}^n \frac 1{k+2} = \sum_{k=1}^n \frac1k - \sum_{k=3}^{n+2} \frac 1{k}\\ = \left(1 + \frac 12 + \sum_{k=3}^n \frac1k\right) - \left(\frac1{n+1} + \frac 1{n+2} + \sum_{k=3}^n \frac1k\right). $$

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  • $\begingroup$ Thank you very much, I now see the solution to the given problem! Are there any tips/tricks for any general given function to see these patterns or does it just take practice? $\endgroup$ – Matthias K. May 16 '20 at 21:26
  • $\begingroup$ @MatthiasK. Practice always helps. That said, with these problems in particular, I think it's important to be able to expand and reorganize sums in a "neat" way; make sure that you can read through and understand your own work. $\endgroup$ – Ben Grossmann May 16 '20 at 21:40
  • $\begingroup$ @MatthiasK. Also, if you are satisfied with this answer, please "accept" it by clicking the checkmark (the $\checkmark$) below the arrows on the left. $\endgroup$ – Ben Grossmann May 16 '20 at 21:42
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$$ \eqalign{ & S = \sum\limits_{k = 1}^n {{1 \over {k\left( {k + 2} \right)}}} \cr & 2S = \sum\limits_{k = 1}^n {{1 \over k} - {1 \over {k + 2}}} = \sum\limits_{k = 1}^n {{1 \over k}} - \sum\limits_{k = 1}^n {{1 \over {k + 2}}} = \cr & = \sum\limits_{k = 1}^n {{1 \over k}} - \sum\limits_{k = 3}^{n + 2} {{1 \over k}} = \sum\limits_{k = 1}^2 {{1 \over k}} - \sum\limits_{k = n + 1}^{n + 2} {{1 \over k}} = \cr & = 1 + {1 \over 2} - {1 \over {n + 1}} - {1 \over {n + 2}} = {3 \over 2} - {{2n + 3} \over {\left( {n + 1} \right)\left( {n + 2} \right)}} \cr} $$

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