Generating Function for Sequence of Sum over k to n of Reciprocal of k by n-k

From ProofWiki
Jump to navigation Jump to search

Theorem

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

$\forall n \in \Z_{\ge 0}: a_n = \ds \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k \paren {n - k} }$


Then $\sequence {a_n}$ has the generating function $\map G z$ such that:

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

and whose terms are:

$a_n = \dfrac {2 H_{n - 1} } n$


Proof

From Product of Generating Functions:

$\map G z = \paren {\map {G_1} z}^2$

where $\map {G_1} z$ is the generating function for $\ds \sum_{k \mathop \ge 1} \dfrac 1 k$.

From Generating Function for Sequence of Reciprocals of Natural Numbers:

$\map {G_1} z = \map \ln {\dfrac 1 {1 - z} }$

Hence:

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


Differentiating $\map G z$ with respect to $z$ gives:

\(\ds \map {G'} z\) \(=\) \(\ds \dfrac \d {\d z} \paren {\paren {\ln \dfrac 1 {1 - z} }^2}\)
\(\ds \) \(=\) \(\ds 2 \ln \dfrac 1 {1 - z} \dfrac \d {\d z} \paren {\ln \dfrac 1 {1 - z} }\) Chain Rule for Derivatives, Derivative of Power
\(\ds \) \(=\) \(\ds 2 \paren {\ln \dfrac 1 {1 - z} } \times \paren {\dfrac 1 {1 - z} }^{-1} \dfrac \d {\d z} \dfrac 1 {1 - z}\) Chain Rule for Derivatives, Derivative of Logarithm Function
\(\ds \) \(=\) \(\ds 2 \paren {\ln \dfrac 1 {1 - z} } \times \paren {\dfrac 1 {1 - z} }^{-1} \times \dfrac {-1} {\paren {1 - z}^2} \map {\dfrac \d {\d z} } {1 - z}\) Chain Rule for Derivatives, Derivative of Power
\(\ds \) \(=\) \(\ds 2 \paren {\ln \dfrac 1 {1 - z} } \times \paren {\dfrac 1 {1 - z} }^{-1} \times \dfrac {-1} {\paren {1 - z}^2} \times \paren {-1}\) Chain Rule for Derivatives, Derivative of Power
\(\ds \) \(=\) \(\ds 2 \paren {\dfrac 1 {1 - z} \ln \dfrac 1 {1 - z} }\) simplifying
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop \ge 0} H_n z^n\) Generating Function for Sequence of Harmonic Numbers


Integrating again with respect to $z$ gives:

\(\ds \map G z\) \(=\) \(\ds \int_0^z \paren {2 \sum_{n \mathop \ge 0} H_n z^n}\)
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop \ge 0} H_n \paren {\int_0^z z^n}\)
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop \ge 0} H_n \dfrac {z^{n + 1} } {n + 1}\) Primitive of Power
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop \ge 1} H_{n - 1} \dfrac {z^n} n\) Translation of Index Variable of Summation

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Generating_Function_for_Sequence_of_Sum_over_k_to_n_of_Reciprocal_of_k_by_n-k&oldid=682337"