letumsnemesislh

2022-07-06

$\underset{n\to \mathrm{\infty }}{lim}n\sqrt{2}\phantom{\rule{thinmathspace}{0ex}}\left(\sqrt{\mathrm{ln}\left(n+1\right)}-\sqrt{\mathrm{ln}n}\right)=0$
But I haven't any ideas how to do it... My calculations shows that this sequence is monotonously decreasing.
I've proved that using inequality $\mathrm{ln}\left(1+{x}^{a}\right)\le ax$ and double-sided theorem.

furniranizq

Expert

$n\sqrt{2}\left(\sqrt{\mathrm{ln}\left(n+1\right)}-\sqrt{\mathrm{ln}n}\right)=\sqrt{2}n\frac{\mathrm{ln}\left(n+1\right)-\mathrm{ln}n}{\sqrt{\mathrm{ln}\left(n+1\right)}+\sqrt{\mathrm{ln}n}}\phantom{\rule{0ex}{0ex}}=\sqrt{2}n\frac{\mathrm{ln}\left(1+\frac{1}{n}\right)}{\sqrt{\mathrm{ln}\left(n+1\right)}+\sqrt{\mathrm{ln}n}}{\sim }_{\mathrm{\infty }}\sqrt{2}n\frac{\frac{1}{n}}{2\sqrt{\mathrm{ln}n}}=\frac{1}{\sqrt{2\mathrm{ln}n}}\to 0$

letumsnemesislh

Expert

Since${a}_{n}:=n\sqrt{2}\left(\sqrt{\mathrm{ln}\left(n+1\right)}-\sqrt{\mathrm{ln}n}\right)=\sqrt{2}n\frac{\mathrm{ln}\left(n+1\right)-\mathrm{ln}n}{\sqrt{\mathrm{ln}\left(n+1\right)}+\sqrt{\mathrm{ln}n}}=\sqrt{2}n\frac{\mathrm{ln}\left(1+\frac{1}{n}\right)}{\sqrt{\mathrm{ln}\left(n+1\right)}+\sqrt{\mathrm{ln}n}},$
we have
${a}_{n}=\sqrt{2}\frac{n\left(\frac{1}{n}-\frac{1}{{n}^{2}}+\frac{1}{{n}^{3}}-\dots \right)}{\sqrt{\mathrm{ln}\left(n+1\right)}+\sqrt{\mathrm{ln}n}}=\sqrt{2}\frac{1-\frac{1}{n}+\frac{1}{{n}^{2}}-\dots }{\sqrt{\mathrm{ln}\left(n+1\right)}+\sqrt{\mathrm{ln}n}}$
it follows that
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=0.$

Do you have a similar question?