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If the sum $\frac{1}{7} + \frac{1\cdot 3}{7\cdot 9} + \frac{1\cdot 3\cdot 5}{7\cdot 9\cdot 11} + ...$ to 20 terms is $\frac{m}{n}$, reduced fraction, then what is $n-4m$?

This is a question I dug up from an old JEE Main paper (India).
It's very intriguing to me because, although it looks simple, I can't seem to find he sum in this question.

The numerator of the nth term of the series seems to be the product of the first n terms of odd numbers. The denominator is similarly made but the sequence starts from 7. I have no idea how to find the sum to n terms in this situation. If it were the sum of odd numbers and not the product, I could have done it easily.

Please explain to me how you find the product of n terms of a sequence
and also the sum to n terms of the given sequence.

I have never been introduced to the product of n terms before, If you could give me a proper intuition for it, it would really make my christmas.

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  • $\begingroup$ consider the ratio of the nth term and the (n+1)th term. After some rearranging, you will be able to telescope the sum. $\endgroup$
    – shyamupa
    Dec 22, 2013 at 22:57
  • $\begingroup$ $m=32$ and $n=129$ $\endgroup$
    – Fred Daniel Kline
    Dec 22, 2013 at 22:58
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    $\begingroup$ @FredKline: aww come on. That's like rubbing the gold in my face even before I figure out the map. There's no point in giving that to me, my friend. Although if you figured out those answers just by the mere sight of this question, then I am in need of borrowing your brain for a while. (PS - I appreciate your effort, though.) $\endgroup$
    – Nick
    Dec 22, 2013 at 23:18

3 Answers 3

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The following is not particularly clever but works.

Note that $$1\cdot 3 \cdot 5 \cdot \cdots \cdot (2N-1) = \frac{(2N)!}{2^N N!}$$

Using this (or just considering ratios of adjacent terms as suggested in comments), each term in the series is $$\frac{120(2n)! (n+3)!}{n!(2n+6)!} = \frac{15}{(2n+1)(2n+3)(2n+5)} = \frac{15}{8}\left(\frac 1 {2n+1} - \frac 2 {2n+3} + \frac 1 {2n+5} \right)$$

Using telescopic sums, one finds the sum of $N$ terms to be $$\frac 1 4 -\frac{15}8 \left( - \frac 1 {2N+3} + \frac 1 {2N+5}\right)$$ and the answer can be calculated.

Edit: As Cameron pointed out in the comments, the telescopic nature of the sum is made clearer by realizing you are summing the terms $$\left(\frac 1 3 - \frac 1 5\right) - \left(\frac 1 5 - \frac 1 7\right),\quad \left(\frac 1 5 - \frac 1 7\right) - \left(\frac 1 7 - \frac 1 9\right),\quad \cdots$$

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    $\begingroup$ "The following is not particularly clever but works." Well, I doubt that there is a much cleverer solution to be found. Usually these types of problems are set up with telescoping in mind. $\endgroup$
    – Eric Naslund
    Dec 22, 2013 at 23:08
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    $\begingroup$ I'm so very sorry but I didn't follow. How did you telescope it? $\endgroup$
    – Nick
    Dec 22, 2013 at 23:13
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    $\begingroup$ @Nick: Would it make it easier to see how it telescopes if you rewrite $$\frac{15}8\left(\frac1{2n+1}-\frac2{2n+3}+\frac1{2n+5}\right)$$ as $$\frac{15}8\left(\frac1{2n+1}-\frac1{2n+3}\right)+\frac{15}8\left(-\frac1{2n+3}+\frac1{2n+5}\right),$$ instead? $\endgroup$
    – Cameron Buie
    Dec 23, 2013 at 3:28
  • $\begingroup$ @EricNaslund - Given the $n-4m = 1$ nature of the answer, I rather suspect there is a better way of doing this. $\endgroup$
    – not all wrong
    Dec 23, 2013 at 15:46
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    $\begingroup$ @Nick I edited the answer in the same spirit as Cameron's comment. $\endgroup$
    – not all wrong
    Dec 23, 2013 at 15:51
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Let's write the last term. What do we notice ? $\dfrac{1\cdot3\cdot5\cdot7\cdot9\cdot11\cdot13\cdot\ldots\cdot39}{7\cdot9\cdot11\cdot\ldots\cdot39\cdot41\cdot43\cdot45}=\dfrac{1\cdot3\cdot5}{41\cdot43\cdot45}$

All terms, starting with the fourth, are of the form $\dfrac{1\cdot3\cdot5}{(2n+1)(2n+3)(2n+5)}$ , which has the

words telescoping series written all over it. :-) Then all that's left to do is add the sum of the first three terms to its result, simplify the fraction, and compute $n-4m$.

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Mathematica code for my comment above. I didn't make it an answer because it is only a programming hack, and doesn't explain the mathematical underpinnings. 

c = Total[
    Join[{(a = 1)/(b = 7)}, 
         Table[(a *= j)/(b *= (j + 6)), {j, 3, 39, 2}]]];
Denominator[c] - 4 Numerator[c]
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    $\begingroup$ to this day, i have never gotten around to installling mathematica and running this. thank you though, this is a much appreciated answer. $\endgroup$
    – Nick
    Aug 18, 2018 at 15:51

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