Sum of Any Infinite Series
Date: 11/19/2000 at 02:00:28 From: Dan Keesing Subject: If an Infinite Series Converges Hi Dr. Math, I just learned about various tests for convergence and the "powerful" Taylor series, but I was disappointed to learn that even if you can determine that a series converges, it's usually impossible to calculate an exact answer for the sum. Except for the rare cases like inf. SUM (1/n^2) = pi^2/6 n=1 by Euler's proof, it would be extremely difficult to find an exact answer for an infinite sum. I am having trouble understanding the idea that if an infinite sum converges, it's not always possible to compute an actual answer - you can only get only a decimal or fractional approximation (using Taylor's theorem for the accuracy). Can you explain to me why it's not usually possible to compute this type of sum? Two examples of sums I'd like to find exactly are: inf. SUM 1/[(2n)!] n=1 and inf. SUM 10^n/[(n+1)*4^(2n+1)] n=1 By the Ratio Test, they both are absolutely convergent, so I'd like to see if it's possible to find an exact answer. Thanks!
Date: 11/21/2000 at 00:00:52 From: Doctor Fenton Subject: Re: If an Infinite Series Converges Dear Dan, You've asked a very interesting question, and in some sense, the question is as much philosophical as mathematical. Part of the answer to your first question, why we can't find a "closed-form" solution to most series, in terms of "familiar" quantities, is connected with the question, "How many irrationals do you know?" Irrational numbers fall into two camps, the algebraic numbers that are roots of polynomials with integer coefficients, such as sqrt(2), sqrt(3), etc., and transcendental numbers such as pi or e. If you know what a countable infinity is, then it will be meaningful to say that there are only countably many finite rational combinations of algebraic numbers, e and its powers e^2, e^3,... , pi and its powers, and rational roots of these, but there are uncountably many real numbers, so "most" real numbers cannot be expressed in terms of pi, e, or algebraic numbers to begin with, no matter how the real number arose. If you're not familiar with countability, there's a good discussion, "Set Theory and Orders of Infinity," here: http://mathforum.org/dr.math/problems/klimkow.4.8.97.html If you search our archives with a keyword like "uncountable" you'll find even more. So, part of the problem is that you (and I) really don't know many irrational numbers by name. There are more real numbers than names, for that matter. The philosophical part of the answer is the question why you feel more comfortable with an expression pi^2/6 than a decimal expansion. After all, you don't really know that many decimal places for pi, and why is writing the answer in terms of a quantity that has to be approximated any better than just giving an expression (the series) that will let you compute any desired number of decimal places. In fact, the known decimal places for pi and e come from similar series expansions. It's similar to the way most students feel more comfortable with degrees than with radians, even though the division of a circle into 360 parts is purely arbitrary, a 3000-year old legacy from the Babylonians, while radians are a "natural" definition of angles in terms of area or arclength. On the other hand, this question does bring up the question of why you should believe in the real numbers in the first place. Even though most people are mystified by i, the complex unit, real numbers are much more mystifying when you dig into them. As for your series, those particular examples can actually be summed. For example, e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ... and 1/e = 1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! + ... so (e + 1/e) = 2(1 + 1/2! + 1/4! + 1/6! + ...) which means your first series is e + 1/e ------- - 1 2 The second series is tougher, but from the exponents on the 10 and 4, it resembles a geometric series. The (n+1) in the denominator can arise from integrating x^n, so I'd begin by trying to write this as inf. SUM (x^n)/(n+1) n=1 If I evaluate this series at x = 10/16, which is 10/4^2, I get your series, except for a factor of 4 in the denominator (we'll worry about that later.) Write 1 inf. x^(n+1) 1 inf. x --- SUM ------- = --- SUM INT t^n dt x n=1 (n+1) x n=1 0 1 x inf. = --- INT [SUM t^n ] dt x 0 n=1 1 x t = --- INT ----- dt x 0 1-t 1 = --- [-x - ln(1-x)] x Now divide this by 4 and evaluate this result at x = 10/16, and you'll have your second series. So, in fact, against the odds, the series you gave just happen to have "familiar" values. If you have further questions, please write again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/
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