ums, Approximations, and Asymptotics Il The area of the bars is at least the area under this curve so we have a lower bound on the n-th harmonic sum 111 (n+1) Remember that n blocks can overhang the edge of a table by Hn block wi we had, say, a million blocks, then this lower bound implies that we could achieve an overhang of at least 1m(.00069 block widths! In fact, since the lower bound of In(n+ 1) grows arbitrarily large, there is no limit on how far the stack can overhang. Of course, this assumes no breeze, defor- mation of the blocks, or gravitational variation as our stack grows thousands of miles high We can get an upper bound on the n-th harmonic number by playing essentially the same game. Now we need a curve that skims just above the bar graph. The curve defined by y=1/a fits the bill 1/2 01234 1n2 The area under this curve is an upper bound on the area of the bar graph and thus on the n-th harmonic sum. But there's a problem: the area under the curve is infinite because y=1/ has a bad asymptote at =0. This is a common problem when bounding sums with integrals and there's a simple solution: take the exact value of the first term(1/1)and then bound the remaining terms(1/2+1/3+..+1/n) with an integral. In this case, weSums, Approximations, and Asymptotics II 5 The area of the bars is at least the area under this curve, so we have a lower bound on the n­th harmonic sum: 1 1 1 1 Hn = + + + . . . + � 1 2 3 n n 1 ≥ dx 0 x + 1 = ln(n + 1) Remember that n blocks can overhang the edge of a table by 1 2Hn block widths. So if we had, say, a million blocks, then this lower bound implies that we could achieve an overhang of at least ln(1, 000, 000 + 1) = 6.907 . . . 2 block widths! In fact, since the lower bound of 1 2 ln(n + 1) grows arbitrarily large, there is no limit on how far the stack can overhang. Of course, this assumes no breeze, defor￾mation of the blocks, or gravitational variation as our stack grows thousands of miles high. We can get an upper bound on the n­th harmonic number by playing essentially the same game. Now we need a curve that skims just above the bar graph. The curve defined by y = 1/x fits the bill. - 6 1 1 1 2 1 3 1 4 1 1/2 y = 1/x 0 1 2 3 4 n − 1 n The area under this curve is an upper bound on the area of the bar graph and thus on the n­th harmonic sum. But there’s a problem: the area under the curve is infinite because y = 1/x has a bad asymptote at x = 0. This is a common problem when bounding sums with integrals and there’s a simple solution: take the exact value of the first term (1/1) and then bound the remaining terms (1/2 + 1/3 + . . . + 1/n) with an integral. In this case, we