
#1
Oct2013, 05:42 AM

P: 246

I'm doing problem section 15 chapter 1 by Boas. I don't want to ask about a particular problem in there but she often gives this kind of instruction, "By computer or tables, find the exact sum of each of the following series." My question is, what kind of table she is referring to? I take it there must be some kind of equivalent table to integral table. I checked Wiki and there's one but it seems to be too general.
I want to find this for example, question in page 41 there's this series. ##\sum_{n=1}^{\infty} \frac{n}{(4n^21)^2}## Is it possible to find the above series using a combination of elementary functions, or is that series unique? I can't find it in the wiki table. Secondly, is it legitimate if I just put the type to WolframAlpha directly to answer the question? (Using computer method) Thank You 



#2
Oct2013, 06:24 AM

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Note that:
[tex]\frac{n}{(4n^{2}1)^{2}}=\frac{n}{(2n+1)^{2}(2n1)^{2}}[/tex] Further partials fractions decomposition, in seeking to rewrite this as the sum: a/(2n+1)^2+b/(2n1)^2 might help you along. You get, twice, that a+b=0, and 4a+4b=1, meaning be=1/8, a=1/8 The two subseries has a nice relation between them. 



#3
Oct2013, 06:58 AM

P: 246

What I did before was to decompose ##((2n1)(2n+1))^1)## first and then try to get the partial sum. After that I square the partial sum and multiply it with the formula for the series of ##n##. Apparently this technique is flawed. Is it okay though as I asked before to use computer because the question allows it? Can all convergent series be nicely decomposed like this? Edit: And oh another thing, why don't you reduce the power of the denominator of the partial fraction? They have a power of 2. Is it okay not to add additional term because all of the denominators have the same power? 



#4
Oct2013, 07:13 AM

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Using Table and Computer to find the sum of a series
You don't need to use terms like a/(2n+1), because it is sufficient with the square terms.
Meaning that if you DO get the correct expressions with a/(2n+1), then these can be recast into the simplified square form I gave. 



#5
Oct2013, 07:15 AM

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"Can all convergent series be nicely decomposed like this?"
If God wills it, perhaps. If he exists at all, something I doubt. 



#6
Oct2013, 07:24 AM

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When Mary Boas wrote her book, software like Mathematica was just a fantasy. When she says to use a computer, I'm sure she meant to write a routine which calculates the terms of the series and produces partial sums until some cutoff criteria was met. Today, we would use a spreadsheet to do such a calculation.
Gradsteyn And Ryzhik include tables of various series in their Table of Integrals: http://atsol.fis.ucv.cl/dariop/sites...ij_Engl._2.pdf This is a handy reference to have. (And yes, your example series is printed in this book along with its sum.) 



#7
Oct2013, 07:48 AM

P: 246

Considering when the Boas book was printed, is the reason why I asked this question. It just seems weird trying to use WA with old books like this. She did say "find the exact sum" though. So how do I best approach this problem then in your opinion to get the most out of it? 



#8
Oct2013, 08:01 AM

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What about most series, which does not appear in those tables?
Do you think they have nice, tidy sums as well? Or is it, perhaps, the other way around: Those series with nicy, tidy sums end up in tables, all the others don't? 



#9
Oct2013, 08:29 AM

P: 246





#10
Oct2013, 08:34 AM

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No, there is no one single, general method to do this. And those formulas you see are ALL analytical expressions.
Computers cannot tell you whether that expression, or one diverging with a trillionth in its value from the expression is correct. at least not yet, they can't. And it should be pi^2/6, not pi/6 And yes: Convergence whether divergence IS the critical issue; it is no point in setting a computer to calculate the sum of a divergent series, is there? But, it might be very useful, once you've established convergence, to let the computer spit out a value YOU know ought to lie "close" to the actual sum neither you or the computer ever will know. 



#11
Oct2013, 08:45 AM

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I think if you examine the particular series from your example, you will see that for the first half dozen terms or so, the denominator gets rather large rather quickly, while the numerator is only increasing incrementally. Therefore, it would be reasonable to conclude on first examination that this particular series should converge rather rapidly.
Now obviously, one could use some clever algebra to manipulate this series into some simpler ones, which could be evaluated directly. But by doing successive partial sums, I think one could deduce the limiting sum rather more quickly. 


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