Find $\lim\limits_{n\to\infty}\frac{a_1+a_2+...+a_n}{1+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt{n}}}$ with $a_1=1$ and $a_{n+1}=\frac{1+a_n}{\sqrt{n+1}}$
Stolz–Cesàro is a way to go, but applied to $S_n=\sum\limits_{k=1}^n a_n$ and $T_n=\sum\limits_{k=1}^n \frac{1}{\sqrt{k}}$, where $T_n$ is strictly monotone and divergent sequence ($T_n >\sqrt{n}$). Then $$\lim\limits_{n\rightarrow\infty}\frac{S_{n+1}-S_n}{T_{n+1}-T_n}= \lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{\frac{1}{\sqrt{n+1}}}= \lim\limits_{n\rightarrow\infty} \left(1+a_n\right)=\\ \lim\limits_{n\rightarrow\infty} \left(1+\frac{1+a_{n-1}}{\sqrt{n}}\right)= \lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n(n-1)}}+\frac{a_{n-2}}{\sqrt{n(n-1)}}\right)=\\ \lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n(n-1)}}+\frac{1}{\sqrt{n(n-1)(n-2)}}+...+\frac{a_1}{\sqrt{n!}}\right)=\\ 1+\lim\limits_{n\rightarrow\infty} \left(\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+...+\frac{1}{\sqrt{(n-1)!}}\right)\right)$$
Now, for $$\lim\limits_{n\rightarrow\infty} \left(\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+...+\frac{1}{\sqrt{(n-1)!}}\right)\right) \tag{1}$$ we have $$0<\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+\frac{1}{\sqrt{(n-1)(n-2)(n-3)}}+...+\frac{1}{\sqrt{(n-1)!}}\right)< \frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+\frac{1}{\sqrt{(n-1)(n-2)}}+...+\frac{1}{\sqrt{(n-1)(n-2)}}\right) =\\\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{n-3}{\sqrt{(n-1)(n-2)}}\right)\rightarrow 0$$
Finally, $(1)$ has $0$ as the limit, $\frac{S_{n+1}-S_n}{T_{n+1}-T_n}$ has $1$ as the limit. The original sequence has $1$ as the limit as well.
Let $b_n=\sqrt{n!}a_n$, then the recursion becomes $$ b_{n+1}=\sqrt{n!}+b_n $$ and we get $$ \begin{align} b_n &=\sum_{k=0}^{n-1}\sqrt{k!}\\ &=\sqrt{(n-1)!}\left(1+\frac1{\sqrt{n-1}}+\frac1{\sqrt{(n-1)(n-2)}}+\dots+\frac1{\sqrt{(n-1)!}}\right)\\ a_n &=\frac1{\sqrt{n}}\left(1+\frac1{\sqrt{n-1}}+\frac1{\sqrt{(n-1)(n-2)}}+\dots+\frac1{\sqrt{(n-1)!}}\right) \end{align} $$ Therefore, the Euler-Maclaurin Sum Formula says $$ \sum_{k=1}^na_k=2\sqrt{n}+\log(n)+C_1+O\!\left(\frac1{\sqrt{n}}\right) $$ Furthermore, $$ \sum_{k=1}^n\frac1{\sqrt{k}}=2\sqrt{n}+C_2+O\!\left(\frac1{\sqrt{n}}\right) $$ Therefore, $$ \lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^na_k}{\displaystyle\sum_{k=1}^n\frac1{\sqrt{k}}}=1 $$