Sequence of General Harmonic Numbers Converges for Index Greater than 1

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 $\dfrac {2^{r - 1} } {2^{r - 1} - 1}$.

$\blacksquare$