Harmonic Number - Calculation

Calculation

An integral representation is given by Euler:

The equality above is obvious by the simple algebraic identity below

An elegant combinatorial expression can be obtained for using the simple integral transform :

\begin{align} H_n &= \int_0^1 \frac{\,\,\, 1 - x^n}{1 - x}\,dx =-\int_1^0\frac{1-(1-u)^n}{u}\,du = \int_0^1\frac{1-(1-u)^n}{u}\,du \\ &= \int_0^1\left\,du \\ &= \sum_{k=1}^n (-1)^{k-1}\binom nk \int_0^1u^{k-1}\,du \\ &= \sum_{k=1}^n(-1)^{k-1}\frac{1}{k}\binom nk . \end{align}

The same representation can be produced by using the third Retkes identity putting and using the fact that .


Hn grows about as fast as the natural logarithm of n. The reason is that the sum is approximated by the integral

whose value is ln(n).

The values of the sequence Hn - ln(n) decrease monotonically towards the limit:

(where γ is the Euler–Mascheroni constant 0.5772156649...), and the corresponding asymptotic expansion as :

where are the Bernoulli numbers.

Read more about this topic:  Harmonic Number

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