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

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?

Answer 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 $$\lim\limits_{n\rightarrow\infty}\frac{S_{n+1}-S_n}{T_{n+1}-T_n}= \lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{\frac{1}{\sqrt{n+1}}}= \lim\limits_{n\rightarrow\infty} \left(1+a_n\right)=\\ \lim\limits_{n\rightarrow\infty} \left(1+\frac{1+a_{n-1}}{\sqrt{n}}\right)= \lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n(n-1)}}+\frac{a_{n-2}}{\sqrt{n(n-1)}}\right)=\\ \lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n(n-1)}}+\frac{1}{\sqrt{n(n-1)(n-2)}}+...+\frac{a_1}{\sqrt{n!}}\right)=\\ 1+\lim\limits_{n\rightarrow\infty} \left(\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+...+\frac{1}{\sqrt{(n-1)!}}\right)\right)$$

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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 $$0<\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+\frac{1}{\sqrt{(n-1)(n-2)(n-3)}}+...+\frac{1}{\sqrt{(n-1)!}}\right)< \frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+\frac{1}{\sqrt{(n-1)(n-2)}}+...+\frac{1}{\sqrt{(n-1)(n-2)}}\right) =\\\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{n-3}{\sqrt{(n-1)(n-2)}}\right)\rightarrow 0$$

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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.

Answer 2

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?

Answer 3

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, $$ \lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^na_k}{\displaystyle\sum_{k=1}^n\frac1{\sqrt{k}}}=1 $$