Relation between harmonic numbers and Stirling numbers of the first kind

Prove that
$$\left[n\atop 2\right] = H_{n-1}\Gamma(n)$$

By induction on \( n\) we have

for \( n=2\)

$$\left[2\atop 2\right] = H_{1}\times\Gamma(1) = 1$$

Assume that

$$\left[k\atop 2\right] = H_{k-1}\Gamma(k)$$

Then by the recurrence relation

$$\left[k+1\atop 2\right] = k\left[k\atop 2\right] + \left[k\atop 1\right]$$

Use the inductive step

$$\left[k+1\atop 2\right] = \Gamma(k+1)H_{k-1} + \Gamma(k) = \Gamma(k+1)\left\{H_k+\frac{1}{k} \right\} = \Gamma(k+1)H_k$$

This entry was posted in Harmonic numbers, Striling numbers of first kind and tagged , , , , , . Bookmark the permalink.

Leave a Reply