analysislimitsreal-analysis
$\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$
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?
$\sum\limits ^{2n}_{k=n}\dfrac{\ln k}{k} \geq (\ln n)(\sum\limits ^{2n}_{k=n}\dfrac{1}{k}) \geq (\ln n)(\sum\limits ^{2n}_{k=n}\dfrac{1}{2n}) \geq \frac{ \ln n}{2}$ which goes to $\infty$
The limit of the ratio in your title is nonzero. Give me a few minutes to type up the following, based on your funtion $g(x) = x^{-1/3}$
if we have $g(x) > 0$ but $g'(x) < 0,$ then $$ \int_a^{b+1} \; g(x) \; dx \; < \; \sum_{j=a}^b \; g(j) \; < \; \int_{a-1}^b \; g(x) \; dx $$
Here is a drawing I made, using the letter $f$ rather than $g$
Well, $g$ is integrable at the origin. Let's try $a=1.$ If that is not satisfactory we can just switch to $a=2$ by putting in some extra terms.
$$ \int_1^{n+1} \; x^{-1/3} \; dx \; < \; \sum_{j=a}^b \; j^{-1/3} \; < \; \int_{0}^n \; x^{-1/3} \; dx $$
an antiderivative of $g$ is $G(x) = \frac{3}{2} x^{2/3}$
$$ \frac{3}{2} \left( (n+1)^{2/3} - 1 \right) \; < \; \sum_{j=1}^n \; j^{-1/3} \; < \; \frac{3}{2} n^{2/3} $$
Good enough. Your denominator is simply $ n^{2/3}.$ We see that $$ L = \frac{3}{2} $$
Best Answer
The limit is $+\infty$.
You made an error in the first step. Applying Stolz-Cesaro, you should rather consider:
\begin{eqnarray*} \frac{\frac 1{\sqrt{n+1}}}{\ln{(n+1)}-\ln n} & = & \frac 1{\sqrt{(n+1)}}\cdot \frac 1{\ln \left(1+\frac 1 n\right)}\\ & = & \frac n{\sqrt{(n+1)}}\cdot \underbrace{\frac{\frac 1n}{\ln \left(1+\frac 1 n\right)}}_{\stackrel{n \to \infty}{\longrightarrow}1} \\ & \stackrel{n \to \infty}{\longrightarrow} & +\infty \end{eqnarray*}