Sequence of General Harmonic Numbers Converges for Index Greater than 1

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Theorem

Let $H_n^{\paren r}$ denote the general harmonic number:

$\ds H_n^{\paren r} = \sum_{k \mathop = 1}^n \frac 1 {k^r}$

for $r \in \R_{>0}$.


Let $r > 1$.

Then as $n \to \infty$, $H_n^{\paren r}$ is convergent with an upper bound of $\dfrac {2^{r - 1} } {2^{r - 1} - 1}$.


Proof

For any $m \in \N$:

\(\ds H_{2^m - 1}^{\paren r}\) \(=\) \(\ds H_{2^{m - 1} - 1}^{\paren r} + \dfrac 1 {\paren {2^{m - 1} }^r} + \dfrac 1 {\paren {2^{m - 1} + 1}^r} + \cdots + \dfrac 1 {\paren {2^{m - 1} + \paren {2^{m - 1} - 1} }^r}\)
\(\ds \) \(<\) \(\ds H_{2^{m - 1} - 1}^{\paren r} + \dfrac 1 {\paren {2^{m - 1} }^r} + \dfrac 1 {\paren {2^{m - 1} }^r} + \cdots + \dfrac 1 {\paren {2^{m - 1} }^r}\) Ordering of Reciprocals
\(\ds \) \(=\) \(\ds H_{2^{m - 1} - 1}^{\paren r} + \dfrac {2^{m - 1} } {2^{\paren {m - 1} r} }\)
\(\ds \) \(=\) \(\ds H_{2^{m - 1} - 1}^{\paren r} + \paren {2^{1 - r} }^{m - 1}\)
\(\ds \) \(<\) \(\ds H_{2^{m - 2} - 1}^{\paren r} + \paren {2^{1 - r} }^{m - 2} + \paren {2^{1 - r} }^{m - 1}\)
\(\ds \) \(<\) \(\ds \dots\)
\(\ds \) \(<\) \(\ds H_{2^0 - 1}^{\paren r} + \paren {2^{1 - r} }^0 + \paren {2^{1 - r} }^1 + \dots + \paren {2^{1 - r} }^{m - 1}\)
\(\ds \) \(=\) \(\ds 0 + \sum_{k \mathop = 0}^{m - 1} \paren {2^{1 - r} }^k\)
\(\ds \) \(=\) \(\ds \dfrac {1 - \paren {2^{1 - r} }^m} {1 - 2^{1 - r} }\) Sum of Geometric Sequence
\(\ds \) \(<\) \(\ds \dfrac 1 {1 - 2^{1 - r} }\) as $0 < 2^{1 - r} < 1$
\(\ds \) \(=\) \(\ds \dfrac {2^{r - 1} } {2^{r - 1} - 1}\)

Since $m$ is arbitrary, every partial sum $H_n^{\paren r}$ is bounded from above by $\dfrac {2^{r - 1} } {2^{r - 1} - 1}$.

By Monotone Convergence Theorem, as $n \to \infty$, $H_n^{\paren r}$ is convergent with an upper bound of $\dfrac {2^{r - 1} } {2^{r - 1} - 1}$.

$\blacksquare$


Sources

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