<s>[INST] A limit of a sequence that yields different answers This is the problem:

$\lim_{n\rightarrow\infty}\frac{1}{n\sqrt{n}}\sum^n_{k=1}\sqrt{k}$

A friend and I tried to solve it using different methods and our results are very different. Our attempts:
First attempt. Using Riemann sums
$\lim_{n\rightarrow\infty}\frac{1}{n\sqrt{n}}\sum^n_{k=1}\sqrt{k}=\lim_{n\rightarrow\infty}\frac{1}{n\sqrt{n}}(\sqrt{1}+...+\sqrt{n})=\lim_{n\rightarrow\infty}\frac{1}{n}(\sqrt{\frac{1}{n}}+...+\sqrt{\frac{n}{n}})=\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1}\sqrt{\frac{k}{n}}$
If we choose the partition $\Pi_n= (0, \frac{1}{n}, \frac{2}{n},..., \frac{n}{n}=1)$ and let $f(\frac{k}{n})=\sqrt{\frac{k}{n}}$:
$\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1}\sqrt{\frac{k}{n}}=\int_0^1\sqrt{x}\ dx=\frac{2}{3}$
Second attempt. Using Stolz-Cèsaro theorem
$\frac{\sqrt{1}+...+\sqrt{n+1}-\sqrt{1}-...-\sqrt{n}}{(n+1)\sqrt{n+1}-n\sqrt{n}}=\frac{1}{(\sqrt{(n+1)^3}-\sqrt{n^3})(\sqrt{n+1}+\sqrt{n})}\rightarrow 0$
We can't figure out why one of these (or both) may be wrong, so any hint or help would be really appreciated.
 [/INST] The second attempt is incorrect. You have made a mistake in your algebra as the first identity written in that part is false. You can try the following
\begin{align*}
\frac{\sqrt{n+1}}{(n+1)\sqrt{n+1}- n\sqrt n} &= \frac{\sqrt{n+1}}{(n+1)\sqrt{n+1}- n\sqrt n} \times \frac{(n+1)\sqrt{n+1}+n\sqrt n }{(n+1)\sqrt{n+1}+n\sqrt n} \\
&=\frac{\sqrt{n+1}\left( (n+1)\sqrt{n+1}+n\sqrt n \right)}{(n+1)^2(n+1)- n^2 n}\\
&= \frac{(n+1)^2 +n\sqrt{n(n+1)} }{(n+1)^3- n^3} \\
&= \frac{(n+1)^2}{(n+1)^3-n^3} + \frac{n\sqrt{n(n+1)} }{(n+1)^3- n^3}
\end{align*}
These two terms both converge to $1/3$ so the limit is $2/3$.
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