The given sum is a Riemann sum for
Integ{n to 2n} 1/x dx
As such, it is an approximation for the latter. Looked at the other way, the integral is an approximation to the sum. As n approaches infinity, the approximation becomes exact.
The antiderivative of 1/x is ln(x), so the definite integral becomes ln(2n) - ln(n). But this is just ln(2), and its limit as n approaches infinity stays ln(2), which, to the shown accuracy, is .693147180559945.
The convergence is shown for chosen values of n:
n sum
1 .5
2 .5833333333333333
3 .6166666666666667
4 .6345238095238096
5 .6456349206349207
6 .6532106782106782
7 .6587051837051836
8 .6628718503718504
9 .6661398242280596
10 .668771403175428
100 .6906534304818244
1000 .6928972430599376
10000 .6931221811849485
100000 .6931446805661912
1000000 .6931469305600098
10000000 .6931471555599048
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Posted by Charlie
on 2004-08-19 10:43:15 |
solution
