Message-ID: <20040201.043923@whim.org>

In article <bvhgn0$be8$1@hercules.btinternet.com>,
"Alex Kendrick" <alex_6136@yahoo.co.uk> wrote:
>can anyone offer any insights on finding the sum to n terms of the harmonic
>series? i've tried everything, but every expression i try boils down to the
>original series. i even ended up considering the logarithm of a variable
>base, which is something i've never even seen before in a maths problem but
>even this approach comes to nothing. i looked on mathworld and they gave a
>formula but it used a function i've heard of but know nothing about, the
>gamma function and a constant. i know the answer is already known, but thats
>not very satisifying, i'd rather derive it myself. By i way, i'm not at
>university yet so i don't have access to the literature, thanks in advance
>for any help. (harmonic series is sigma 1/n)

If you want to understand why the partial sum of the harmonic series is
the Euler-Mascheroni constant plus the digamma function, see

    http://www.whim.org/nebula/math/partharm.html

The Euler-Mascheroni constant is approximately .5772156649...  If you
want to see one method of computing this constant, see

    http://www.whim.org/nebula/math/eulergamma.html

I have used this algorithm to compute over 10000 digits of gamma.

If all you need is an approximation of the partial harmonic sum, you
can use the Euler-Maclaurin Sum Formula to get the asymptotic series

     n
    --- 1
    >   -
    --- k
    k=1

    = log(n) + gamma

       1     1        1        1        1        1
    + -- - ----- + ------ - ------ + ------ - ------- + ...
      2n   12n^2   120n^4   252n^6   240n^8   132n^10

The error in this approximation should be less than 1/(47n^12).

For more about the Euler-Maclaurin Sum Formula see

    http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html

or

    http://www.whim.org/nebula/math/eulermac.html

Rob Johnson