Infinite Series of 1/nDate: 01/13/98 at 08:56:17 From: Michael Theemling Subject: Sum of an infinite series of 1/n when n=1,2,3... This is a question I encountered in calculus a long time ago while I was still a young math student. I understand the answer is divergence, or the sum is infinity, but do not conceptually understand why that is so, especially since the terms eventually go to 0. 1 +1/2+1/3+1/4+... For n>1 and n<1 it is relatively easy to comprehend, but not this one. I saw a proof before but never really got it. I just want to add your site is outstanding. As a student (now graduated) in mathematics, I found your information clear and your knowledge in-depth. Date: 01/26/98 at 10:59:21 From: Doctor Joe Subject: Re: Sum of an infinite series of 1/n when n=1,2,3... Dear Michael, The divergence of 1 + 1/2 + 1/3 + 1/4 + ... is famous (or rather infamous!). The proof requires some geometrical interpretation and a minimal knowledge of integral calculus. Let f(x) = 1/x. The next thing you need to do is to sketch this reciprocal function over the interval (1,infinity). Next, construct "boxes" that cover the region bounded by the curve y=f(x), the y=1 line, the x-axis, and the y=n+1 line in the following way: On each integer interval a rectangle is constructed by taking the height to be the maximum value over that interval. For example, on the interval [1,2], the maximum value of the function y = 1/x is 1. y 1 | ____ | | | | | | | | | 1/2 | | |____ | | | | 1/3 | | | | | | | | | | | | ----------------------- x 0 1 2 3 ... If we form the sum of the areas of the first n rectangles, then the value is: 1 + 1/2 + 1/3 + 1/4 + ... + 1/n Since the curve is concave downwards, the sum of areas of the n+1 rectangles is clearly greater than the area under the curve f(x) = 1/x on the interval [1,n+1]. So, we have the following inequality: n+1 Int | 1/x < 1 + 1/2 + 1/3 + 1/4 + ... + 1/n 1 for each n, positive integer. Evaluating the integral on the left, we have ln(n+1). Suppose that the sum 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... converges to, say, a real number S. Then, lim ln (n+1) < or equal to S. n->infinity Note that the sequence {ln(n+1)} diverges to infinity. Thus, we have arrived at a contradiction, and therefore our supposition is false. Hence, 1 + 1/2 + 1/3 + ... diverges. (Classical analysts call this series the harmonic series.) -Doctor Joe, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 04/03/00 at 08:36:42 From: Dr. Schwa Subject: harmonic series To prove that the sum diverges, you can also use a comparison test to a series that obviously diverges. Group the terms together: (1) + (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + ... + 1/16) + ... taking twice as many terms each time. This is clearly greater than (1) + (1/2) + (1/4 + 1/4) + (1/8 + 1/8+ 1/8 + 1/8) + (1/16 + ... + 1/16) + ... since we replaced each term with something smaller or equal. Then, each of the parenthesized groups is 1/2, so the original harmonic series is greater than 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + ... Since there are infinitely many 1/2's, the series diverges. -Dr. Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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