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

I've proved that using inequality $\ln(1+x^a) \le ax$ and double-sided theorem.
 A: $$n\sqrt{2}\left(\sqrt{\ln(n+1)}-\sqrt{\ln n}\right)=\sqrt{2}n\frac{\ln(n+1)-\ln n}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}\\=\sqrt{2}n\frac{\ln\left(1+\frac{1}{n}\right)}{\sqrt{\ln(n+1)}+\sqrt{\ln n}}\sim_\infty\sqrt 2 n\frac{\frac{1}{n}}{2\sqrt{\ln n}}=\frac{1}{\sqrt{2\ln n}}\to0$$
A: 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
$$
\lim_{n\to \infty}a_n=0.
$$
A: \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: $$n\sqrt 2 (\sqrt{\ln(n+1)}-\sqrt{\ln n})\to 0.$$
A: 
this is my try for this limit please
$$ \lim_{x\rightarrow \infty} n \sqrt{2} (\sqrt{\ln(n+1)} - \sqrt{\ln(n)}) $$
$$ = \lim_{x\rightarrow \infty} n \sqrt{2}
 \frac{\ln(n+1) - \ln(n)}{\sqrt{\ln(n+1)} + \sqrt{\ln(n)}} $$
$$ = \lim_{x\rightarrow \infty} \frac{\sqrt{2}}{\sqrt{\ln(n+1)} + \sqrt{\ln(n)}} \cdot
 \frac{\ln\left( 1 + \frac{1}{n} \right)}{\frac{1}{n}} = 0\cdot 1 = 0 $$
