Sum of Sequence of Harmonic Numbers/Proof 3
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Theorem
- $\ds \sum_{k \mathop = 1}^n H_k = \paren {n + 1} H_n - n$
where $H_k$ denotes the $k$th harmonic number.
Proof
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 $\sequence {a_n}$.
From Generating Function for Sequence of Harmonic Numbers:
- $\map G z = \dfrac 1 {1 - z} \map \ln {\dfrac 1 {1 - z} }$
Then:
\(\ds \map {G'} z\) | \(=\) | \(\ds \dfrac 1 {\paren {1 - z}^2} \map \ln {\dfrac 1 {1 - z} } + \dfrac 1 {\paren {1 - z}^2}\) | Derivative of Generating Function for Sequence of Harmonic Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {1 - z} \map G z + \dfrac 1 {\paren {1 - z}^2}\) |
From Generating Function for Sequence of Partial Sums of Series, $\dfrac 1 {1 - z} \map G z$ is the generating function for $\sequence {b_n}$ where:
- $b_n = \ds \sum_{k \mathop = 0}^n H_k$
and so:
- $\dfrac 1 {\paren {1 - z}^2} \map \ln {\dfrac 1 {1 - z} } = \ds \sum_{n \mathop \ge 0} \paren {\sum_{k \mathop = 0}^n H_k} z^n$
From Generating Function for Natural Numbers:
- $\dfrac 1 {\paren {1 - z}^2} = \ds \sum_{n \mathop \ge 0} \paren {n + 1} z^n$
That is:
- $\map {G'} z = \ds \sum_{n \mathop \ge 0} \paren {\sum_{k \mathop = 0}^n H_k + \paren {n + 1} } z^n$
Now we have:
\(\ds \map {G'} z\) | \(=\) | \(\ds \map {\dfrac \d {\d z} } {\sum_{n \mathop \ge 0} H_n z^n}\) | Derivative of Generating Function for Sequence of Harmonic Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop \ge 0} n H_n z^{n - 1}\) | Derivative of Power |
Equating coefficients of $z^n$ in these two expressions for $\map {G'} z$:
- $\ds \sum_{k \mathop = 0}^n H_k + \paren {n + 1} = \paren {n + 1} H_{n + 1}$
\(\ds \sum_{k \mathop = 0}^n H_k + \paren {n + 1}\) | \(=\) | \(\ds \paren {n + 1} H_{n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {n + 1} H_n + \paren {n + 1} \paren {\dfrac 1 {n + 1} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{k \mathop = 1}^n H_k + n\) | \(=\) | \(\ds \paren {n + 1} H_n\) | $H_k = 0$ when $k = 0$, and simplification |
The result follows.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.9$: Generating Functions: Exercise $3$