Comparison of the Harmonic Series We’re going to compare: � ∞ � 1 n n=1 ∞ dx , using Riemann sums to show that the series diverges. The same x 1 sort of reasoning is applicable to the previous two examples. We’ll use a Riemann sum to calculate the area under the graph of y = x1 using Δx = 1. We also get to choose whether to compute the upper Riemann sum or lower; we’ll do both. to y=⁄x 1 ½ 1 ½ 1 2 3 Figure 1: Upper Riemann sum; y = ⅓ 1 x (not to scale). The upper Riemann sum is the sum of the areas of the rectangles indicated in Figure 1. If we stop at the rectangle just before N we see that the area under �N the curve from 1 to N is 0 dx x and that that’s less than the sum of the areas of the rectangles: � 1 N dx 1 1 1 < 1 + + + ··· + . x 2 3 N −1 (There are only N − 1 rectangles here because the distance between 1 and N is N − 1.) If Sn = 1 + 12 + 13 + · · · + N1 , then because there’s one more term in the sum we have: � N dx 1 1 1 < 1 + + + ··· + < SN . x 2 3 N −1 1 Thus we know that the value of the integral is less than the value of SN . This will allow us to prove conclusively that the series diverges. � 1 N dx x 1 < SN N < SN ln N − ln 1 < SN ln N − 0 < SN ln N < SN ln x|1 As N goes to inﬁnity, ln N goes to inﬁnity and so SN must also go to inﬁnity. By deﬁnition, ∞ � 1 = lim SN , N →∞ n n=1 �∞ so the series n=1 n1 diverges. y=⁄x ¼ ½ 1 ⅓ 2 3 Figure 2: Lower Riemann sum; y = 4 1 x (not to scale). Now we’ll use the lower Riemann sum to see that SN goes to inﬁnity at �N the same rate as 1 dx x . Since the sum of the areas of the rectangles shown in Figure 2 is less than the area under the curve, we know that: � 1 Recall that �N 1 dx x N dx 1 1 1 > + + ··· + = SN − 1. x 2 3 N = ln N , so we get: � N dx x 1 < (ln N ) + 1. SN − 1 < SN Combining this with the result from the upper Riemann sum, we conclude: ln N < SN < (ln N ) + 1. The value of SN is hard to calculate exactly, but now we know that it’s between ln N and (ln N ) + 1, which we can compute. 2