This series looks like it is geometric or sort of, but the rate between them does not seem constant.

$$S_{n}= \frac{1}{3\times 6}+\frac{1}{6\times 9}+\frac{1}{9\times 12}+ \cdots + \frac{1}{300\times 303}$$

All i can see is the second number in the denominator jumps to the next term in the series but this situation makes it impossible to establish a common factor. Therefore, Is it possible to determine the number of terms and the sum using a simple algorithm?.

There is also a side question which I don't know. In the example from above the final term is $\frac{1}{300\times 303}$ but what if the series goes to the infinity. Is this series convergent or not?. How can this be proven?.


Although there is a method to approach these kind of situations mentioned here (What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots +\frac{1}{n(n+1)}$), it does not address the fact on how to obtain a recursive formula, as it is part of the question itself. Therefore is there a way on how to reach to that formula in the denominator and solve the problem?. If the method of solving this involves partial fractions I would appreciate somebody could include a revision of this part in the answer so I can understand how it makes a link with the problem from above.


marked as duplicate by Simply Beautiful Art, Donald Splutterwit, Namaste algebra-precalculus Oct 1 '17 at 23:04

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  • $\begingroup$ Hint: Partial fractions. $\endgroup$ – Simply Beautiful Art Oct 1 '17 at 22:34
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    $\begingroup$ In the displayed equation there's an $n$ on the left side but no $n$ on the right side. Should that $S_n$ be $S_{100}?$ $\endgroup$ – bof Oct 1 '17 at 22:37
  • $\begingroup$ Mind I ask why you are using "$\doteq$" instead of simply "$=$"? $\endgroup$ – Simply Beautiful Art Oct 1 '17 at 22:49
  • $\begingroup$ @SimplyBeautifulArt Sorry, I didn't notice this when used the Latex editor, now its fixed. $\endgroup$ – Chris Steinbeck Bell Oct 1 '17 at 22:54
  • $\begingroup$ "it does not address the fact on how to obtain a recursive formula" I tried searching for the word "recursive" and only got one pop-up, which was is the quoted line. $\endgroup$ – Simply Beautiful Art Oct 1 '17 at 22:56

Define $$a_n = \frac{1}{3n*3(n+1)}$$ Then $$S_n=a_1+a_2+...+a_n$$ Note that $$a_n=\frac{1}{9} \frac{1}{n(n+1)}=\frac{1}{9}(\frac{1}{n}-\frac{1}{n+1}) $$ It follows that $$S_n=\frac{1}{9}\left( \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{9}\left( 1-\frac{1}{n+1}\right)$$

  • $\begingroup$ Mind if you sharing a bit on how did you obtained $a_{n}$? $\endgroup$ – Chris Steinbeck Bell Oct 1 '17 at 23:33
  • $\begingroup$ @ChrisSteinbeckBell How do you define the $n$th fraction in your sum? $\endgroup$ – Simply Beautiful Art Oct 1 '17 at 23:41
  • $\begingroup$ @ChrisSteinbeckBell Just observe the definition of each term in your sum, and write out the general formula for the nth term. $\endgroup$ – Jiaqi Li Oct 2 '17 at 0:53
  • $\begingroup$ @JiaqiLi It can take some time to get to the general formula but once its done the rest seems simple. Can this reasoning be extended to other examples or is it specific for this kind of series?. $\endgroup$ – Chris Steinbeck Bell Oct 2 '17 at 2:32
  • $\begingroup$ @ChrisSteinbeckBell It is a specific technique for this kind of series, in which each term can be split into two terms, and they cancel out when added together. We call this "term fission" in China. $\endgroup$ – Jiaqi Li Oct 2 '17 at 18:19

\begin{eqnarray*} S_{100}= \frac{1}{3 \times 6} + \frac{1}{6 \times 9} + \cdots + \frac{1}{300 \times 303} \\ S_{100}=\frac{1}{9} \left( \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{100 \times 101} \right)\\ S_{100}= \frac{1}{9} \sum_{i=1}^{100} \frac{1}{i(i+1)} \end{eqnarray*} This sum can be summed using partial fractions \begin{eqnarray*} \frac{1}{i(i+1)}= \frac{1}{i} -\frac{1}{i+1} \\ S_{100}=\frac{1}{9} \left( 1 - \frac{1}{ 101} \right)\\ \end{eqnarray*} The sum to infinty is \begin{eqnarray*} S_{\infty}=\frac{1}{9} \\ \end{eqnarray*}


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