Question Number 73378 by mind is power last updated on 10/Nov/19
$${Hello}\:,{i}\:{shar}\:{withe}\:{you}\:{nice}\:{problem}\: \\ $$$${show}\:{that}\:\forall{k}\in\mathbb{N}^{\ast} \:\exists{n}\in\mathbb{N}\:{such}\:{that} \\ $$$${k}\leqslant\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{j}}<{k}+\mathrm{1} \\ $$$${have}\:{a}\:{very}\:{Nice}\:{day} \\ $$$$ \\ $$
Commented by MJS last updated on 11/Nov/19
$${H}_{{n}} =\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{j}};\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{H}_{{n}} =+\infty \\ $$$$\forall{n}\in\mathbb{N}:\begin{cases}{{H}_{{n}+\mathrm{1}} −{H}_{{n}} <\mathrm{1}}\\{{H}_{{n}+\mathrm{2}} −{H}_{{n}+\mathrm{1}} <{H}_{{n}+\mathrm{1}} −{H}_{{n}} }\end{cases} \\ $$$${k}=\mathrm{1} \\ $$$$\mathrm{1}\leqslant{H}_{\mathrm{1}} <{H}_{\mathrm{2}} <{H}_{\mathrm{3}} <\mathrm{2} \\ $$$$\Rightarrow\:\forall{k}\in\mathbb{N}^{\bigstar} \exists{n}\in\mathbb{N}:{k}\leqslant\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{j}}<{k}+\mathrm{1} \\ $$
Commented by mind is power last updated on 11/Nov/19
$${nice}\:\:\:{solution}\: \\ $$
Commented by MJS last updated on 11/Nov/19
$$\mathrm{thank}\:\mathrm{you} \\ $$