Sum of Sequence of Harmonic Numbers/Proof 1

From ProofWiki
Jump to navigation Jump to search

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

\(\ds \sum_{k \mathop = 1}^n H_k\) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {\sum_{j \mathop = 1}^k \frac 1 j}\) 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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Sequence_of_Harmonic_Numbers/Proof_1&oldid=664746"