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 $a_{n+1}=\frac{1+a_n}{\sqrt{n+1}}$ 
Let $(a_n)_{n\ge1}, a_1=1, a_{n+1}=\frac{1+a_n}{\sqrt{n+1}}$.
  Find $$\lim_{n\to\infty} \frac{a_1+a_2+\cdots+a_n}{1+\frac{1}{\sqrt2}+\cdots+\frac{1}{\sqrt{n}}}$$

These is my try:
I intercalated the limit like that
$$L=\lim_{n\to\infty} \frac{a_1+a_2+\cdots+a_n}{\sqrt{n+1}}\frac{\sqrt{n+1}}{1+\frac{1}{\sqrt2}+\cdots+\frac{1}{\sqrt{n}}}$$.
The second term  of the limit tends to 2.
The first one, after Cesaro-Stols, become:
$$\lim_{n\to\infty}a_{n+1}(\sqrt{n+1}+\sqrt{n+2})$$
I tried to intercalate the term $a_n$ between 2 terms in function of n, just like $a_n<\frac{1}{\sqrt{n}}$ or something like that to use the sandwich theorem. Any ideas of this kind of terms? Or other ideas for the problem?
 A: 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.
A: 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,
$$
\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^na_k}{\displaystyle\sum_{k=1}^n\frac1{\sqrt{k}}}=1
$$
A: First, I think you flipped one of your limits as
$$\lim_{n\to\infty}\frac{\sqrt{n+1}}{1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}}=\frac{1}{2}$$
Now, let us find a bound on $a_n$ as $n$ goes to infinity. It is easy enough to see that $a_1=1$, $a_2=1.41421$, $a_3=1.39385$, $a_4=1.19692$, and $a_5=0.982494<1$. By induction, assume $a_n<1$ (with $n\geq 5$). Then we have
$$a_{n+1}=\frac{1+a_n}{\sqrt{n+1}}<\frac{2}{\sqrt{n+1}}\leq\frac{2}{\sqrt{5+1}}<1.$$
Thus, $a_{n+1}<1$ for $n\geq 5$ and we may conclude that the sequence is bounded above by $2$. Can you finish it from here?
