Find $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$ $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$
Can we apply  Stolz-Cesaro?
$\lim\limits_{n \to \infty}\frac {\frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n}}} {{\ln(n+1)-\ln(n)}}$ =
$\lim\limits_{n \to \infty}\frac{\sqrt n - \sqrt {n+1}}{\sqrt n \sqrt{n+1}\ln(1+\frac{1}{n})}$ = $\lim\limits_{n \to \infty}\frac{1-\sqrt{1+\frac{1}{n}}}{\sqrt{n+1}\ln(1+\frac{1}{n})}$
What can I do from here?

Option:
$n/√n < 1/√1+1/√2....+1/√n;$
$\dfrac{√n}{2 \log (√n)} < \dfrac{1/√1+1/√2+...1/√n}{\log n};$
$(1/2) \lim_{n \rightarrow \infty} \dfrac{√n}{\log √n} = $?
