Computational Insights

We begin with a straightforward observation: each term in the harmonic sum can be visualized as a rectangle with a width of 1 and a height of 1/i. Hence, H(n) can be interpreted as the cumulative area of specific rectangles, represented by the yellow area in the following illustration.

Notice how the rectangles are arranged so that the graph of the function f(x) = 1/x for x ≥ 1 intersects the top left corner of each rectangle, remaining within the yellow area. This indicates that H(n) equals the total area of the first n rectangles shown. From this, we can deduce that the yellow area is at least as expansive as the area under the red line, meaning it is also greater than the area between the graph of f(x) = 1/x and the x-axis:

This already provides a good lower estimate, particularly indicating that as n increases, H(n) becomes unbounded, confirming that the harmonic series diverges:

However, we seek a more precise approximation. By adjusting the rectangles to ignore the first term and shifting them one unit left, we can observe that the function f(x) = 1/x encompasses at least as much area as the remaining rectangles. This is visually represented below.

From the above figure, we conclude:

This leads us to establish that

Specifically, we find that

These bounds indicate that for any n, if we compute ln(n) instead of H(n), our error will be no greater than 1.

Refining the Approximation

With much more extensive analysis (beyond the scope of this article), it can be shown that the error term in the approximation converges to a constant that is actually smaller than 1. This constant is known as the Euler-Mascheroni constant, formally defined as:

The first 50 digits of this constant are:

γ = 0.57721566490153286060651209008240243104215933593992.

Using this constant, it can be demonstrated that the n-th Harmonic Number is effectively represented as: