$\lim_{n \to \infty} n\sqrt 2\, \big(\sqrt{\ln(n+1)}-\sqrt{\ln n}\big) = 0$

Question

I'm trying to prove that $$ \lim_{n \to \infty} n\sqrt{2}\,\left(\sqrt{\ln(n+1)}-\sqrt{\ln n}\right) = 0 $$ But I haven't any ideas how to do it... My calculations shows that this sequence is monotonously decreasing.

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I've proved that using inequality $\ln(1+x^a) \le ax$ and double-sided theorem.

Answer 1

Since $$ a_n:=n\sqrt{2}\left(\sqrt{\ln(n+1)}-\sqrt{\ln n}\right)=\sqrt{2}n\frac{\ln(n+1)-\ln n}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}=\sqrt{2}n\frac{\ln\left(1+\frac{1}{n}\right)}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}, $$ we have $$ a_n=\sqrt{2}\frac{n\left(\frac{1}{n}-\frac{1}{n^2}+\frac{1}{n^3}-\ldots\right)}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}=\sqrt{2}\frac{1-\frac{1}{n}+\frac{1}{n^2}-\ldots}{\sqrt{\ln(n+1)}+\sqrt{\ln n}} $$ it follows that $$ \lim_{n\to \infty}a_n=0. $$

Answer 2

$$n\sqrt{2}\left(\sqrt{\ln(n+1)}-\sqrt{\ln n}\right)=\sqrt{2}n\frac{\ln(n+1)-\ln n}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}\\=\sqrt{2}n\frac{\ln\left(1+\frac{1}{n}\right)}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}\sim_\infty\sqrt 2 n\frac{\frac{1}{n}}{2\sqrt{\ln n}}=\frac{1}{\sqrt{2\ln n}}\to0$$

Answer 3

this is my try for this limit please

$$ \lim_{x\rightarrow \infty} n \sqrt{2} (\sqrt{\ln(n+1)} - \sqrt{\ln(n)}) $$ $$ = \lim_{x\rightarrow \infty} n \sqrt{2} \frac{\ln(n+1) - \ln(n)}{\sqrt{\ln(n+1)} + \sqrt{\ln(n)}} $$ $$ = \lim_{x\rightarrow \infty} \frac{\sqrt{2}}{\sqrt{\ln(n+1)} + \sqrt{\ln(n)}} \cdot \frac{\ln\left( 1 + \frac{1}{n} \right)}{\frac{1}{n}} = 0\cdot 1 = 0 $$

Answer 4

\begin{align} n\sqrt 2 (\sqrt{\ln(n+1)}-\sqrt{\ln n}) &=n\sqrt 2 \frac{\ln(n+1)-\ln n}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}=n\sqrt 2\frac{\ln(1+\frac{1}{n})}{\sqrt{\ln(n+1)}+\sqrt{\ln n}} \\ &=\sqrt 2\frac{\ln(1+\frac{1}{n})^n}{\sqrt{\ln(n+1)}+\sqrt{\ln n}} \end{align} Now $$ \ln\left(1+\frac{1}{n}\right)^n\to \ln\mathrm{e}=1, $$ while $$ \frac{1}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}\to 0. $$ Thus finally: $$n\sqrt 2 (\sqrt{\ln(n+1)}-\sqrt{\ln n})\to 0.$$