# Sequence of General Harmonic Numbers Converges for Index Greater than 1

## 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

$\blacksquare$