Derivative of Generating Function for Sequence of Harmonic Numbers

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Theorem

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

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

where $H_n$ denotes the $n$th harmonic number.

Let $\map G z$ be the generating function for $\sequence {a_n}$:

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

from Generating Function for Sequence of Harmonic Numbers.

Then the derivative of $\map G z$ with respect to $z$ is given by:

$\map {G'} z = \dfrac 1 {\paren {1 - z}^2} \map \ln {\dfrac 1 {1 - z} } + \dfrac 1 {\paren {1 - z}^2}$


Proof

\(\ds \map {G'} z\) \(=\) \(\ds \map {\dfrac \d {\d z} } {\dfrac 1 {1 - z} \map \ln {\dfrac 1 {1 - z} } }\)
\(\ds \) \(=\) \(\ds \dfrac 1 {1 - z} \map {\dfrac \d {\d z} } {\map \ln {\dfrac 1 {1 - z} } } + \map \ln {\dfrac 1 {1 - z} } \map {\dfrac \d {\d z} } {\dfrac 1 {1 - z} }\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac 1 {1 - z} \paren {\paren {-1} \dfrac {-1} {\paren {1 - z}^2} \paren {\dfrac 1 {1 - z} }^{-1} } + \map \ln {\dfrac 1 {1 - z} } \paren {-1} \paren {\dfrac {-1} {\paren {1 - z}^2} }\) Derivative of Logarithm Function, Chain Rule for Derivatives, Derivative of Power
\(\ds \) \(=\) \(\ds \dfrac 1 {\paren {1 - z}^2} \map \ln {\dfrac 1 {1 - z} } + \dfrac 1 {\paren {1 - z}^2}\) simplifying

$\blacksquare$


Sources

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