All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Summing inverses (Posted on 2004-08-19)
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 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

 Search: Search body:
Forums (0)