<s>[INST] Need to prove inequality $\sum\limits_{k=0}^n \frac{1}{(n+k)} \ge \frac{2}{3}$ Prove that for $n \geq 1$:
$$\sum\limits_{k=0}^n \frac{1}{(n+k)} \ge \frac{2}{3}$$
I have tried math induction but that didn't work. Although I'm pretty sure that the solving can be done by induction
 [/INST] We can replace the lower bound $\frac{2}{3}$ with $\frac{15}{22}$. Since $A_1=\frac{3}{2}$ and for any $n\geq 1$
$$A_n=\sum_{k=0}^{n}\frac{1}{n+k}=H_{2n}-H_{n-1}$$
satisfies:
$$ A_{n}-A_{n+1} = \frac{6n+4}{2n(2n+1)(2n+2)} \leq\frac{3}{4}\left(\frac{1}{n-1/12}-\frac{1}{n+11/12}\right)$$
we have:
$$\begin{eqnarray*} A_n &=& A_1 + (A_2-A_1) + (A_3-A_2) + \ldots + (A_n-A_{n-1})\\ &\geq& \frac{3}{2}-\frac{3}{4}\sum_{j=1}^{n-1}\left(\frac{1}{j-1/12}-\frac{1}{j+11/12}\right)\geq \frac{3}{2}-\frac{3}{4}\cdot\frac{12}{11}=\frac{15}{22}>\frac{2}{3}.\end{eqnarray*}$$
To avoid non-trivial inequalities, we can just show that the sequence $\{A_n\}_{n\in\mathbb{N}^*}$ is decreasing and prove that:
$$ \lim_{n\to +\infty} A_n = \int_{0}^{1}\frac{dx}{1+x} = \log 2$$
by a Riemann-sums argument. This implies $A_n\geq \log 2>\frac{9}{13}$.
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