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I have been manipulating a certain series for several hours without finding any pattern. Hence I am wondering what some of the better strategies are to find patterns and thus an explicit formula for a series. Among the things I have tried so far are:

  • looking for a common difference between terms

  • looking for a common ratio between terms

  • reversing the order of the terms and summing them up, to check whether the result will be the same for all terms

  • bringing the terms to a common denominator and looking for a obvious pattern in the numerator

I had no luck with any of these and others. The series btw. is $\sum_{k = 1}^n\frac{k - 1}{k(k + 1)(k + 2)}$

This is oen of the things I tried:

$\begin{align*} S_n & = \frac{0}{6} + \frac{1}{24} + \frac{2}{60} + \frac{3}{120} + \frac{4}{210} + \frac{6}{336}\\ & = \frac{0}{6} + \frac{1}{24} + \frac{1}{30} + \frac{1}{40} + \frac{1}{52,5} + \frac{1}{56}\\ & = \frac{0}{1680} + \frac{70}{1680} + \frac{56}{1680} + \frac{42}{1680} + \frac{32}{1680} + \frac{30}{1680}\\ a_n - a_{n+1} : & -\frac{70}{1680}; \frac{14}{1680}; \frac{14}{1680}; \frac{10}{1680}; \frac{8}{1680}; \end{align*}$

The differences between the terms get ever smaller and the sum approaches $.25$, but any internal pattern remains hidden after the things I tried. So, are there a number of useful methods to uncover patterns in series?

user3578468
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    I don't have an answer for you, but I want to applaud you for explaining what you've tried, and what results you've gotten or not gotten. I'm going to leave it to someone else to actually answer, but if you split the numerator, then one of the two terms is answered by this question: http://math.stackexchange.com/questions/1934746/determine-sum-of-the-series-sum-k-1-infty-frac1nn1n2?rq=1. In general, "partial fractions" is the secret sauce for problems of this kind. – John Hughes Dec 11 '16 at 20:07

1 Answers1

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Hints:

  • $\require{cancel}\;\sum_{k = 1}^n\frac{k - 1}{k(k + 1)(k + 2)} = \sum_{k = 1}^n\frac{\cancel{k}}{\cancel{k}(k + 1)(k + 2)} - \sum_{k = 1}^n\frac{1}{k(k + 1)(k + 2)}$

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

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

dxiv
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    @user3578468: Note the strategic element of this hint is Telescoping series. (+1) – Markus Scheuer Dec 11 '16 at 21:11
  • @dxiv Thanks, I so far have reconstructed all steps, but I struggle with the last one, by partial fraction decomposition of $\frac{1}{k(k+1)(k+2)}$ into $\frac{x}{k} + \frac{y}{k + 1} + \frac{z}{k+2}$ I arrived at $\frac{1}{k(k+1)(k+2)} = \frac{.5}{k} + \frac{-1}{k + 1} + \frac{.5}{k+2}$ which is in fact the same, but how did you arrive at the two fractions rather than my three? – user3578468 Dec 12 '16 at 01:28
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    @user3578468 That one was, lacking a better word, "by inspection" (it's not a full fraction decomposition):

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

    It is similar in a way to:

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

     – dxiv Dec 12 '16 at 01:34
  • Ah, I see. Thanks. – user3578468 Dec 12 '16 at 01:37