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 July 18th, 2008, 12:52 PM #1 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 2,802 Thanks: 32 Need a book recommendation Lately, my work has involved formulae that are in series form, and I'm not very well equipped to deal with them. I was able to solve the degenerate case of one particular series after spending a lot of time on it. But, now I have to find a more general solution to this function, and need a good resource to educate myself. I know I could just post the problem here and let everyone have a go at it, but I'd like the challenge of trying it myself first. Can anyone recommend a good book or books that covers the various techniques involved in finding sums of series? In particular, I'd like to find something that can give me some insight in finding expressions for partial sums.
 August 7th, 2008, 01:03 AM #2 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 2,802 Thanks: 32 Re: Need a book recommendation Hmm. after re-reading what I posted, I'm not surprised there are no suggestions. My original post looks very confusing. Anyway, it looks like my original problem has now been solved, more or less. And I found in the library, an english translation of Leonard Euler's book on infinite series, which is excellent reading, so I'll digest that for a while. In the meantime, for those of you who like a good puzzle, here's a series that I was trying to evaluate as part of my main problem. I did figure it out, but I'm interested in finding out whether anyone else can find a simpler way to get the answer than what I had to do. Here is the series: 1/12 + 1/60 + 1/168 + 1/360 + 1/660 + ... The answer is fairly simple, but can you find a more elegant way to get to it than what I did?
 August 7th, 2008, 04:57 AM #3 Member   Joined: Jul 2008 From: Minnesota, USA Posts: 52 Thanks: 0 Re: Need a book recommendation Woah, I can't even figure out the rule for the sequence yet, let alone the sum. Am I missing something easy?
 August 7th, 2008, 05:59 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Need a book recommendation Ditto. The numbers are smooth, but the factors don't accumulate.
 August 7th, 2008, 04:24 PM #5 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 2,802 Thanks: 32 Re: Need a book recommendation Actually, that is nearly all the information that I was given, other than the fact that the sum to 4 decimal places is 0.1137. Now you have all the information that I had. I managed to figure out the pattern of the sequence from just those 5 terms, and then derive the sum analytically. For a bit of background, this was part of the function used to generate a table of values in an old electrical engineering text (circa 1946, before the days of computers). I was trying to determine the actual formula for the terms so that I could calculate the values to higher accuracy. In fact, it is a degenerate case (for n=1) of the series: $\frac{1}{12n^2} + \frac{1}{60n^4} + \frac{1}{168n^6} + \frac{1}{360n^8} + \frac{1}{660n^{10}} + \cdots$ It converges very quickly for n>1, but extremely slowly for n=1.
 August 8th, 2008, 05:24 AM #6 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Need a book recommendation Hi First we notice that the terms in the sequence are of the form $\frac{1}{12\cdot1},\qquad\frac{1}{12\cdot5},\qquad \frac{1}{12\cdot14},\qquad\frac{1}{12\cdot30}, \qquad\frac{1}{12\cdot55},\qquad\dots$ The numbers 1,5,14,30,55,... are the sum of all squares from 1 to n (i.e. 1,1+4,1+4+9, etc.) and you can check that the formula for these numbers is $\sum_{j=1}^n j^2=\frac16n(n+1)(2n+1).$ Therefore, the nth term in the sequence is given by the formula $\frac{1}{2n(n+1)(2n+1)}.$ We can split this expression into the sum of three fractions with denominators n, (n+1) and (2n+1); this gives us the expression $\frac{1}{2n(n+1)(2n+1)}=\frac1{2n}+\frac1{2(n+1)}-\frac2{2n+1}.$ Therefore, the total sum of the terms is $\sum_{n=1}^\infty\left(\frac1{2n}+\frac1{2(n+1)}-\frac2{2n+1}\right).$ We now do a little trick and change the index on the second term from n+1 to n. However, we need to subtract 1/2 so that the sum still stays the same. To see why, write the sum as given in the expression above out term by term as follows: $\qquad\,\frac12+\frac14-\frac23\\[10pt] +\frac14+\frac16-\frac25\\[10pt] +\frac16+\frac18-\frac27\\ \dots$ We shift the middle column down by one row: $\qquad\,\frac12\qquad\qquad-\frac23\\[10pt] +\frac14+\frac14-\frac25\\[10pt] +\frac16+\frac16-\frac27\\ \dots$ then add and subtract 1/2: $\qquad\,\frac12+\frac12-\frac23-\frac12\\[10pt] +\frac14+\frac14-\frac25\\[10pt] +\frac16+\frac16-\frac27\\ \dots$ to see that we can indeed write the sum as $\qquad-\frac12+\sum_{n=1}^\infty\left(\frac1{2n}+\frac1{2 n}-\frac2{2n+1}\right)\\[10pt] =-\frac12+2\sum_{n=1}^\infty\left(\frac1{2n}-\frac1{2n+1}\right)\\[10pt] =-\frac12+2\sum_{m=2}^\infty\frac{(-1)^m}{m}.$ Since we know (from the power series of the log function) that $\sum_{m=1}^\infty\frac{(-1)^m}{m}=\log 2,$ we can write the sum as $-\frac12+2(1-\log2)=\frac32-2\log2.$ Is that how you did it? This probably isn't the most elegant way of doing it!
 August 9th, 2008, 01:52 AM #7 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 2,802 Thanks: 32 Re: Need a book recommendation Hi Mattpi, Thanks for looking at this. My method was a bit different. I factored the coefficients and then used partial fractions to split each term into three sub-terms the same way you did, but then I noted that the subterms were similar to the terms of the harmonic series except that they were shifted and/or multiplied by a constant. So, I wrote an expression for the partial sum of the series in terms of harmonic numbers: The sum of the $\frac{1}{2n}$ terms is $\frac12H_n$ The sum of the $\frac{1}{2(n+1)}$ terms is $\frac12(H_{n+1}-1)$ The sum of the $\frac{-2}{2n+1}$terms is $-2(H_{2(n+1)} -H_{n+1}-1)$ So the partial sum expression for the series is: $S_n= \frac12H_n + \frac12(H_{n+1}-1) -2(H_{2(n+1)} -H_{n+1}-1)$ $= \frac12H_n +\frac32H_{n+1} -2H_{2(n+1)} + \frac32$ Since the harmonic numbers are given by: $H_n=Log(n)+\frac1n+\gamma_n$ I substituted this into the partial sum formula to get: $S_n= \frac12(Log(n)+\frac1n+\gamma_n)+\frac32(Log(n+1)+ \frac{1}{n+1}+\gamma_{n+1})-2(Log(2n+2)+\frac{1}{2n+2}+\gamma_{2n+2} +\frac32$ $= [\frac12(Log(n)+\frac32(Log(n+1)-2(Log(2n+2)] + [\frac1n+\frac{1}{2n+2}] + [\gamma_n+\gamma_{n+1}-2\gamma_{2n+1}] + \frac32$ As $n\to\infty$ the 1/n terms in the second set of brackets go to zero, and the gamma terms inside the 3rd set of brackets converge to Euler's constant and cancel each other out. That leaves the log terms inside the first set of brackets: $S= \lim_{n\to\infty}S_n = \lim_{n\to\infty}\frac12(Log(n)+\frac32(Log(n+1)-2(Log(2n+2) +\frac32$ $= \lim_{n\to\infty} Log( \frac{sqrt{n}(n+1)sqrt{n+1}}{(2n+2)^2})+\frac32$ which converges to $Log(\frac14)+\frac32$ which is equivalent to what you came up with. Thanks for posting your method. It's good to see how others approach the problem. EDIT: BTW, I hadn't noticed that the coefficients/12 were the sum of squares between 1 and n. I had simply factored the coefficients down to primes and then tried recombining them in different ways until I noticed the following pattern: 12 = 2 x 2x 3 60 = 3 x 4 x 5 168 = 4 x 6 x 7 360 = 5 x 8 x 9 660 = 6 x 10 x 11 The formula for the coefficients was then obviously: (n+1) x (2n) x (2n+1)

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