Sum of Sequence of Harmonic Numbers
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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 1
\(\ds \sum_{k \mathop = 1}^n H_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {\sum_{j \mathop = 1}^k \frac 1 j}\) | Definition of Harmonic Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \paren {\sum_{k \mathop = j}^n \frac 1 j}\) | Summation of i from 1 to n of Summation of j from 1 to i | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \paren {\paren {1 + 1 + \cdots + 1} + \paren {\dfrac 1 2 + \dfrac 1 2 + \cdots + \dfrac 1 2} + \cdots + \paren {\dfrac 1 n} }\) | $n$ copies of $1$, $\paren {n - 1}$ copies of $\dfrac 1 2$, $\paren {n - 2}$ copies of $\dfrac 1 3$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \paren {n + 1 - j} \times \frac 1 j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \frac {n + 1 - j} j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \frac {n + 1} j - \sum_{j \mathop = 1}^n \frac j j\) | Linear Combination of Convergent Series | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {n + 1} H_n - n\) | Linear Combination of Convergent Series and Definition of Harmonic Numbers |
$\blacksquare$
Proof 2
From Sum over k to n of k Choose m by kth Harmonic Number:
- $\ds \sum_{k \mathop = 1}^n \binom k m H_k = \binom {n + 1} {m + 1} \paren {H_{n + 1} - \frac 1 {m + 1} }$
Setting $m = 0$:
\(\ds \sum_{k \mathop = 1}^n \binom k 0 H_k\) | \(=\) | \(\ds \binom {n + 1} {0 + 1} \paren {H_{n + 1} - \frac 1 {0 + 1} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{j \mathop = 1}^n H_k\) | \(=\) | \(\ds \paren {n + 1} \paren {H_{n + 1} - 1}\) | Binomial Coefficient with $0$, Binomial Coefficient with $1$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \paren {n + 1} \paren {H_n + \frac 1 {n + 1} - 1}\) | Definition of Harmonic Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {n + 1} H_n + \paren {n + 1} \frac 1 {n + 1} - \paren {n + 1} 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {n + 1} H_n + 1 - \paren {n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {n + 1} H_n - n\) |
$\blacksquare$
Proof 3
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.
$\blacksquare$