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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?

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  • $\begingroup$ This is a problem from Gazeta Matematica No 9, a Romanian magazine. It is no longer an ongoing problem, but it may still be chosen for the District stage of the Romanian Mathematics Olympiad. Whether this should be closed or not is moot. $\endgroup$ – Alexdanut Mar 10 at 10:39
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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.

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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?

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  • $\begingroup$ I see what you did here, but i can't put beetwen 2 termens to use sandwich theorem. $\endgroup$ – Radu Mar 9 at 21:06
  • $\begingroup$ If $a_n$ is bounded by $2$, then it must go to zero as $0<a_{n+1}=\frac{1+a_n}{\sqrt{n+1}}<\frac{3}{\sqrt{n+1}}$. $\endgroup$ – Nick Guerrero Mar 9 at 21:13
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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, $$ \lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^na_k}{\displaystyle\sum_{k=1}^n\frac1{\sqrt{k}}}=1 $$

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  • $\begingroup$ $C_1\doteq1.109767433111392255013424942$ and $C_2=\zeta(1/2)\doteq-1.460354508809586812889499153$. $\endgroup$ – robjohn Mar 17 at 1:56

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