Generating Function for Sequence of Reciprocals of Natural Numbers

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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:

$\displaystyle \map G z = \sum_{n \mathop = 1}^\infty z^n = \frac 1 {1 - z}$

By Integral of Generating Function:

\(\displaystyle \int_0^z \map G t \rd t\) \(=\) \(\displaystyle \sum_{k \mathop \ge 1} \dfrac {z^k} k\)
\(\displaystyle \) \(=\) \(\displaystyle z + \dfrac {z^2} 2 + \dfrac {z^3} 3 + \dfrac {z^4} 4 + \cdots\)

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:

\(\displaystyle \int_0^z \map G t \rd t\) \(=\) \(\displaystyle \int_0^z \dfrac 1 {1 - t} \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \map \ln {\dfrac 1 {1 - z} }\)


The result follows from the definition of generating function.

$\blacksquare$


Sources