Sequence of General Harmonic Numbers Converges for Index Greater than 1

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Theorem

Let $H_n^{\left({r}\right)}$ denote the general harmonic number:

$\displaystyle H_n^{\left({r}\right)} = \sum_{k \mathop = 1}^n \frac 1 {k^r}$

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


Let $r > 1$.

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


Proof

\(\displaystyle H_{2^m}^{\left({r}\right)}\) \(=\) \(\displaystyle H_{2^{m - 1} }^{\left({r}\right)} + \dfrac 1 {\left({2^{m - 1} + 1}\right)^r} + \dfrac 1 {\left({2^{m - 1} + 2}\right)^r} + \cdots + \dfrac 1 {\left({2^{m - 1} + 2^m}\right)^r}\)
\(\displaystyle \) \(\le\) \(\displaystyle H_{2^{m - 1} }^{\left({r}\right)} + \dfrac 1 {\left({2^{m - 1} }\right)^r} + \dfrac 1 {\left({2^{m - 1} }\right)^r} + \cdots + \dfrac 1 {\left({2^{m - 1} }\right)^r}\)
\(\displaystyle \) \(=\) \(\displaystyle H_{2^{m - 1} }^{\left({r}\right)} + \dfrac {2^m} {2^{\left({m - 1}\right) r} }\)
\(\displaystyle \) \(=\) \(\displaystyle H_{2^{m - 1} }^{\left({r}\right)} + \dfrac 1 {2^{m \left({r - 1}\right)} }\)
\(\displaystyle \implies \ \ \) \(\displaystyle H_{2^{m + 1} }^{\left({r}\right)}\) \(\le\) \(\displaystyle \sum_{0 \mathop \le k \mathop < m} \dfrac {2^k} {2^{k r} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {2^{r - 1} } {2^{r - 1} - 1}\) Sum of Geometric Progression

$\blacksquare$



Sources

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