Let H0 = 0 and Hn = 1/1 + 1/2 + ... + 1/n.
Show that, for n > 0, Hn = 1 + (H0 + H1 + ... + Hn−1)/n.
We will use mathematical induction on n, the harmonic sum subscript.
The induction will consist of two steps:
The basis is straightforward.
We simply verify that H1 = 1 + H0/1, which is indeed the case.
For the induction step, we assume the inductive hypothesis: Hk = 1 + (H0 + ... + Hk−1)/k.
It will be convenient below to write this as kHk = k + H0 + H1 + ... + Hk−1.
| By definition, Hk+1 | = Hk + 1/(k+1) |
| = [(k + 1)Hk + 1]/(k + 1) | |
| = (kHk + Hk + 1)/(k + 1) |
| By the inductive hypothesis, Hk+1 | = (k + H0 + H1 + ... + Hk−1 + Hk + 1)/(k + 1) |
| = 1 + (H0 + H1 + ... + Hk)/(k + 1) |
This concludes the induction step: we have proved that the proposition holds for k + 1.
Therefore, for n > 0, Hn = 1 + (H0 + H1 + ... + Hn−1)/n.)
Source:
The harmonic sum and a surprising inductive property (document since taken down), by Doug Hensley
Spot the fallacious step in this "proof" by induction that all people in Canada are the same age!