# Generating Function for Sequence of Reciprocals of Natural Numbers

## Theorem

Let $\sequence {a_n}$ be the sequence defined as:

$\forall n \in \N_{> 0}: a_n = n$

That is:

$\sequence {a_n} = 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots$

Then the generating function for $\sequence {a_n}$ is given as:

$\map G z = \map \ln {\dfrac 1 {1 - z} }$

## Proof

Take the sequence:

$S_n = 1, 1, 1, \ldots$

From Generating Function for Constant Sequence, this has the generating function:

$\ds \map G z = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$
 $\ds \int_0^z \map G t \rd t$ $=$ $\ds \sum_{k \mathop \ge 1} \dfrac {z^k} k$ $\ds$ $=$ $\ds z + \dfrac {z^2} 2 + \dfrac {z^3} 3 + \dfrac {z^4} 4 + \cdots$

which is the power series whose coefficients are $\sequence {a_n}$.

But:

$\map G z = \dfrac 1 {1 - z}$

and so by Primitive of Reciprocal and the Integration by Substitution:

 $\ds \int_0^z \map G t \rd t$ $=$ $\ds \int_0^z \dfrac 1 {1 - t} \rd t$ $\ds$ $=$ $\ds \map \ln {\dfrac 1 {1 - z} }$

The result follows from the definition of generating function.

$\blacksquare$