where n is > 0 and an integer k(n) = ln(n) $$h(n) =\sum_{i=1}^n \frac{1}{i}$$
prove h(n) > k(n) ive tried and considered that h(n) is a divergent series so when n approaches infinity, im not sure how to prove that h(n) is bigger becuase the series sum doesn't converge to a value. I think k(n) also approaches infinity.
On
Expanding on arugula's comments, we can write $$ \ln(n)=\int_1^n\frac{dt}{t} = \sum_{k=1}^{n-1} \int_{k}^{k+1} \frac{dt}{t} \le \sum_{k=1}^{n-1}\frac{1}{k}\int_{k}^{k+1}dt = \sum_{k=1}^{n-1}\frac{1}{k} = h(n)-\frac{1}{n} < h(n)$$ where we used the fact that $\frac{1}{k}\ge \frac{1}{t}$ for $t\in [k,k+1].$
Let us consider, $x∈[1,\infty)$, and $\varphi:x\rightarrow \frac{1}{\lfloor x \rfloor}$.
$\displaystyle\int_1^{n+1}\varphi(x)dx=\sum_{k=1}^n\int_k^{k+1}f(x)dx=\sum_{k=1}^nk$
Let us consider, $\psi:x\rightarrow\frac{1}{x}$.
$\lfloor x\rfloor≤x$, hence $\frac{1}{x}≤\frac{1}{\lfloor x \rfloor}$, which means $\psi(x)\leq \varphi(x)$.
$\displaystyle\int_1^{n+1}\psi(x)dx=\int_1^{n+1}\frac{1}{x}dx=\ln(n+1)−\ln1=\ln(n+1)$
Let us consider, $f:x↦\begin{cases}1 & if & x\leq 2 \\ \frac{1}{x-1} & if & x>2\end{cases}$.
$x−1\leq\lfloor x \rfloor$, hence $\frac{1}{\lfloor x \rfloor}\leq\frac{1}{x−1}$, which means $\varphi(x)\leq f(x)$ for $x>2$; for $x\in[1,2],\varphi(x)=f(x)=1$.
$\displaystyle\int_1^{n+1}f(x)dx=1+\ln n$
Now since, $\psi(x)≤\varphi(x)≤f(x)$,
Hence, $\ln(n+1)\leq 1+\frac{1}{2}+\dots+\frac{1}{n}≤1+\ln n$