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I found this question on an old exam paper - UK GCE A-Level (1972) - equivalent to university entrance level in most countries I believe. The method may be "standard" but has left me stumped. Maybe I am missing something obvious. Can someone give me a hint rather than a full worked solution?

Question

Calculate: $$\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{5\cdot 6\cdot 7}+\dfrac{1}{9\cdot 10\cdot 11}+\cdots$$

What do I notice?

It is an infinite series, so one of Geometric, Maclaurin, Taylor Series might be useful. The sum converges because each term is less than geometric series with ratio (0.5).

The terms are formed from "truncated" factorials (my expression)

So the series can be rewritten

$$\frac{0!}{3!}+\frac{4!}{7!}+\frac{8!}{11!}+\cdots$$

There are three successive positive integers in the denominators of each term in the original series and the multiples of 4 are missing from the denominators.

The integers "within" the factorials in the numerator and denominator are (arithmetically) increasing by 4.

Because it is an infinite series I can't hope to "group" the terms by finding common multiples.

So I get stuck.

Then I cheat and put: $\displaystyle\sum \frac{(4k-4)!}{(4k-1)!}$ into Wolfram Alpha.

The answer $\frac{\ln(2)}{4}$, pops out. So I feel an approach to solution might have something to do with the Maclaurin expansion of $\ln(1+x)$ but I can't get anywhere with this.

Any hints would be gratefully received.

Thanks,

Clive

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    $\begingroup$ Hint You may write $$ \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{2}{n+1} + \frac{1}{n+2} \right). $$ Can you see that telescoping takes place when we sum this? $\endgroup$ Feb 16, 2017 at 20:49
  • $\begingroup$ So split each term using partial fractions and find that lots of terms cancel / simplify and I should be able to spot the series ln(2) in there? CRL $\endgroup$ Feb 16, 2017 at 20:51
  • $\begingroup$ I apologize that I misread the series. PFD will help you, but there will be no cancellation. You may use the following trick $$ \frac{1}{(4n+1)(4n+2)(4n+3)} = \frac{1}{2} \int_{0}^{1} (x^{4n} - 2x^{4n+1} + x^{4n+2}) \, dx $$ to convert the infinite sum into an integral. $\endgroup$ Feb 16, 2017 at 21:02
  • $\begingroup$ @SangchulLee I was in the process of writing this way forward and didn't see your comment until after I posted the solution. -Mark $\endgroup$
    – Mark Viola
    Feb 16, 2017 at 21:10
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    $\begingroup$ @SangchulLee Ah! Thank you my friend. Much appreciative!! $\endgroup$
    – Mark Viola
    Feb 16, 2017 at 21:11

2 Answers 2

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We can write the general term of the series as

$$\frac{1}{(4k+1)(4k+2)(4k+3)}=\frac{1}{2}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)$$

Then, noting that $\int_0^1 x^{4k}\,dx=\frac{1}{4k+1}$, we have

$$\begin{align} \sum_{n=0}^\infty\frac{1}{2}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)&=\sum_{n=0}^\infty\frac12 \int_0^1 (x^{4k}-2x^{4k+1}+x^{4k+2})\,dx\\\\ &=\frac12 \int_0^1 \frac{(1-2x+x^2)}{1-x^4}\,dx\\\\ &=-\frac12\int_0^1 \frac{x-1}{(x^2+1)(x+1)}\,dx \end{align}$$

Can you finish now?

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  • $\begingroup$ Ok - I will have a go. If I have this right, the "trick" is to "reverse engineer" the partial fractions to be the results of definite integrals of a particular polynomial then "merge" the sum and the integral into the definite integral of a rational function ?(lot of words there) Thanks, CRL $\endgroup$ Feb 16, 2017 at 21:13
  • $\begingroup$ I think you've got it! Well done. It is a standard trick. ;-)) $\endgroup$
    – Mark Viola
    Feb 16, 2017 at 21:14
  • $\begingroup$ I am experiencing some relief that a "trick" is involved. Thanks to Zev for cleaning up my post with Latex. Thanks for the clear and incredibly swift replies Sangchul and Dr MV. CRL $\endgroup$ Feb 16, 2017 at 21:17
  • $\begingroup$ You're welcome. My pleasure. -Mark $\endgroup$
    – Mark Viola
    Feb 16, 2017 at 21:19
  • $\begingroup$ Best vote? There's a new voting option that let's me select my favorite answer?! $\endgroup$ Feb 18, 2017 at 1:34
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I have a suspicion that the following method would be more like the one expected of the candidates for this exam.

First we decompose into partial fractions, so, as given already, $$S=\frac 12\sum_{r=0}^{\infty}\left(\frac{1}{4k+1}-\frac{2}{4k+2}+\frac{1}{4k+3}\right)$$

Then we start by writing this out explicitly, so that $$2S=\left(\frac 11-\frac 22+\frac 13\right)+\left(\frac 15-\frac 26+\frac 17\right)+\left(\frac 19-\frac{2}{10}+\frac{1}{11}\right)+...$$

Then we systematically add in and subtract terms, so $$2S=\left(\frac 11-\frac 12+\frac 13-\frac 14\right)+\color{red}{\left(-\frac 12+\frac 14\right)}+\left(\frac 15-\frac 16+\frac 17-\frac 18\right)+\color{red}{\left(-\frac 16+\frac 18\right)}+\left(\frac 19-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\right)+\color{red}{\left(-\frac{1}{10}+\frac{1}{12}\right)}+...$$

So $$2S=\ln 2-\color {red}{\frac 12\ln 2}$$

Then $$S=\frac 14\ln 2$$

I don't think the integration method as shown by @Dr. MV was expected to be known by those students...

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  • $\begingroup$ Thank you David.Very neat. $\endgroup$ Feb 17, 2017 at 10:52
  • $\begingroup$ You are welcome $\endgroup$ Feb 17, 2017 at 11:55
  • $\begingroup$ Creatively done :-) $\endgroup$ Feb 18, 2017 at 1:33
  • $\begingroup$ @DavidQuinn The OP mentioned Maclaurin series and typically integration is introduced well before series. $\endgroup$
    – Mark Viola
    Feb 18, 2017 at 2:12
  • $\begingroup$ @Dr.MV. Certainly those students would have known integration, but probably not the method you used to sum a series. The trick you use is very good of course (which is why I upvoted your answer) but I don't recall it ever being taught for A level. Instead candidates were supposed to know standard series such as those based on $ln(1+x)$ and $e^x$ and manipulate them, using reindexing and so on, to sum other not so obvious series. I may be wrong, of course. $\endgroup$ Feb 18, 2017 at 10:45

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