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 An extended harmony (Posted on 2011-11-01)
How many members of the harmonic series 1+1/2+1/3+1/4+ …+1/n are needed to add up close to 10 , without going over it?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 A lower bound and a stab at a solution | Comment 2 of 5 |
I decided to take a look at wikipedia.  It reminded me of the integral test.

The sum of the first n terms > integral of 1/x dx from 1 to k+1 = ln(k+1)

Which implies the solution to ln(k+1)=10 is a lower bound.
k = e^10 -1 = 22025.46
***edit this is an upper bound***

Below that on the wikipedia entry is the rate of divergence:
The sum of the first n terms = ln(k) + EMC + error
EMC is the Euler-Masceroni Constant which is about .5772156649
and the error term is about 1/(2k)

Using the lower bound for the k in the error term gives k=e^(10-EMC-1/(2*22025.46) = 12366.87

I am pretty sure that the error term is small enough to imply 12366 is the solution to the problem.  (12367 would exceed 10)

Edited on November 1, 2011, 10:40 am
 Posted by Jer on 2011-11-01 10:37:54

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