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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?.

Edit:

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.

Michael Hardy
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Chris Steinbeck Bell
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  • Hint: Partial fractions. – Simply Beautiful Art Oct 01 '17 at 22:34
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    In the displayed equation there's an $n$ on the left side but no $n$ on the right side. Should that $S_n$ be $S_{100}?$ – bof Oct 01 '17 at 22:37
  • Mind I ask why you are using "$\doteq$" instead of simply "$=$"? – Simply Beautiful Art Oct 01 '17 at 22:49
  • @SimplyBeautifulArt Sorry, I didn't notice this when used the Latex editor, now its fixed. – Chris Steinbeck Bell Oct 01 '17 at 22:54
  • "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. – Simply Beautiful Art Oct 01 '17 at 22:56
  • Also, what's not recursive about telescoping? – Simply Beautiful Art Oct 01 '17 at 22:56
  • @SimplyBeautifulArt To obtain the formula in the denominator requires the kind of observation you can get by practicing a lot or perhaps by having this learned somewhere. What if I do not have this ability yet?. Therefore I stress the part on how to deduce this without referring to telescoping series?. – Chris Steinbeck Bell Oct 01 '17 at 23:04
  • And how is that not answered in the mentioned duplicate? – Simply Beautiful Art Oct 01 '17 at 23:20
  • @SimplyBeautifulArt First the most voted answer begins straight with a formula not a way to derive it. Also, the denominator is the same as in the question (from the other post). While the other uses integrals. Which is not intuitive for the casual learner. For such reason i think it is not covered in the other question. Now, to get this 'telescoping effect' (the way i call it) you need to factorize a term. Where should I start looking to find that term. This is probably what is needed to understand the solution other than just making it seem that it came like a rabbit out of a hat. – Chris Steinbeck Bell Oct 01 '17 at 23:32
  • The first answer does not, as far as I can tell, give a straight away formula (not that this is important to your question). Nor does it use telescoping arguments. Indeed, I think it is quite a natural solution for your problem. The derivation is left up to the reader, and likely, induction was intended. I do not see how one could get much more recursive than induction itself (both relying fundamentally on the principle of well ordering) – Simply Beautiful Art Oct 01 '17 at 23:39
  • Also note that you still haven't answered bof's clarification question, which was over an hour ago. – Simply Beautiful Art Oct 01 '17 at 23:43
  • @SimplyBeautifulArt Sorry, but i do believe it does give a straight formula.(the one which has 19 votes). The one you are referring indeed it does not mention telescoping properties while it is a natural solution it will work for someone who (again) has a lot of practice on identifying a pattern (regardless of its simplicity), isn't there a way without resorting to heuristic approach?. I believe it does, but probably requires memorization of some equation as the second guy provided. Gee, i never thought to be discussing this on my birthday. :( – Chris Steinbeck Bell Oct 01 '17 at 23:48
  • @ChrisSteinbeckBell At this point, you are basically saying "Can we prove $A$ without using $B$ or $C$ or $D$ ..." Honestly, plugging in a few numbers is one of the things I believe you should do before tackling problems such as these, it honestly helps/can't hurt. And happy birthday. – Simply Beautiful Art Oct 01 '17 at 23:51
  • @bof How do I find the n=100, just factorizing and that's it?. Can this be extended for all cases involving this kind of situation?. If its yes, then I will take it as a way to go. This is the reason why I wrote it as n and not 100. – Chris Steinbeck Bell Oct 01 '17 at 23:52
  • @SimplyBeautifulArt I differ with what you say. Maybe it would be much better to show alternatives in the answer that can be used and which are the requirements to be needed in order to use them (and not be concice). I take your advise about plugging numbers but engaging problems it is a decision that its up to me to make and thanks for your greetings btw. – Chris Steinbeck Bell Oct 02 '17 at 00:03
  • Well, I'd like to say that I'm very good at sequences and series (ofc, this is up to you to decide). From my experience, if you've ever written an answer yourself, including multiple alternatives in one answer is a) difficult to do and b) discouraging for other answerers. – Simply Beautiful Art Oct 02 '17 at 00:13
  • I believe that if someone feels very sure of himself then he should not be carried away by what others may think. Of course, I humbly have no experience in answering questions as I am a beginner. I can only think that someone with more flight hours makes those decisions considering in advance how to make the experience more friendly for others. Even so I am pleased although it has been marked as duplicate I have been given alternatives on what to research and study more. – Chris Steinbeck Bell Oct 02 '17 at 02:42

2 Answers2

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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 $$S_n=\frac{1}{9}\left( \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{9}\left( 1-\frac{1}{n+1}\right)$$

Jiaqi Li
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  • Mind if you sharing a bit on how did you obtained $a_{n}$? – Chris Steinbeck Bell Oct 01 '17 at 23:33
  • @ChrisSteinbeckBell How do you define the $n$th fraction in your sum? – Simply Beautiful Art Oct 01 '17 at 23:41
  • @ChrisSteinbeckBell Just observe the definition of each term in your sum, and write out the general formula for the nth term. – Jiaqi Li Oct 02 '17 at 00:53
  • @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?. – Chris Steinbeck Bell Oct 02 '17 at 02:32
  • @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. – Jiaqi Li Oct 02 '17 at 18:19
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\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*}

Donald Splutterwit
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