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Summing inverses (Posted on 2004-08-19) Difficulty: 3 of 5
What's the limit, as n→∞, of 1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n)?

See The Solution Submitted by Federico Kereki    
Rating: 4.0000 (5 votes)

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Solution solution | Comment 3 of 18 |

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
 

  Posted by Charlie on 2004-08-19 10:43:15
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