TITLE: 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 $a_{n+1}=\frac{1+a_n}{\sqrt{n+1}}$
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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,
$$
\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^na_k}{\displaystyle\sum_{k=1}^n\frac1{\sqrt{k}}}=1
$$