perplexus.info :: Calculus : Summing inverses

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)

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A short solution

| Comment 6 of 18 |

The sum is H(2n)-H(n), H(j) being 1+1/2+1/3+...+1/j.

The limit of H(j) is ln(j) plus a certain constant, called the "Euler-Mascheroni" constanT.

So, H(2n)-H(n) reduces to ln(2n)-ln(n)=ln(2).

Posted by Oskar on 2004-08-19 11:28:19

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