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Let $(a_n)_{\geq 2}$ with $a_2 > 0$ and $$a_{n+1}+1=(na_n+1)^{\frac{1}{n}}$$ Prove that $\displaystyle\frac{a_{n+1}}{a_n}\to 1$ and $na_n \to 0$, eventually using the inequality $$\ln(1+x)>\frac{x}{x+1}, \: \forall x>0$$

I managed to prove that $a_n \to 0$. Using Bernoulli's inequality we get that $(a_n)$ is decreasing and it is obviously bounded by 0. If it were convergent to $l \neq 0$, we yould have $$\lim_{n \to \infty}\frac{\ln(na_n+1)}{n}=\lim_{n \to \infty}\frac{\ln(na_n+1)}{na_n}\cdot a_n=0\cdot l=0$$ since $na_n \to \infty$ and so $\lim_{n \to \infty}(na_n+1)^{\frac{1}{n}}=1$ hence $$l+1=\lim_{n \to \infty}(a_{n+1}+1)=\lim_{n \to \infty}(na_n+1)^{\frac{1}{n}}=1$$ which is a contradiction, so $l=0$.

This is where I got stuck. I don't know how to approach either of these two limits, but I'm pretty sure that their hint with the $\ln(x+1)$ inequality must be used somehow.

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$a_{n+1}+1\leq a_{n}+1$, implying that $\frac{a_{n+1}}{a_n}\leq 1$, is just a consequence of the AM-GM inequality.
On the other hand $n a_n\to 0$ implies that for any $n$ large enough $$ a_{n+1}+1=\exp\left(\frac{1}{n}\log(1+na_n)\right)>\exp\left(\frac{a_n}{na_n+1}\right)>\frac{a_n}{na_n+1}+1 $$ hence $\frac{a_{n+1}}{a_n}\geq 1-\varepsilon$ for any fixed $\varepsilon>0$.
In order to show $na_n\to 0$, you may notice that $a_n\to 0$ implies $$ \log(a_{n+1}+1)= a_{n+1}-O(a_{n+1}^2) = \frac{1}{n}\log(1+na_n) $$ and $n a_{n+1} -O(na_{n+1}^2) = \log(1+na_n)$ implies $na_n\to 0$.

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  • $\begingroup$ Thank you! But how to prove that $na_n \to 0$? $\endgroup$ – AndrewC Mar 21 '18 at 10:04
  • $\begingroup$ @AndrewC: answer updated. $\endgroup$ – Jack D'Aurizio Mar 21 '18 at 12:39
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$\textbf{A bit long but maybe correct!}$

First we can show that $a_n>-1$ for all $n$. For $n=2$ it is true since by assumption $a_2>0$. Suppose there is some $m$ such that for the first time $a_m\leqslant-1$ then $$0<1+(m-1)a_{m-1}=(a_m+1)^{m-1}=(-1)^{m-1}(-1-a_m)^{m-1}$$ which would not hold in case $m$ was even. If instead $m$ was odd then the above inequality would hold. However in the next two iterations we get $$0\geqslant1-m\geqslant1+ma_m=(1+a_{m+1})^m\Rightarrow a_{m+1}\leqslant-1$$ implying $$0>1-(m+1)\geqslant1+(m+1)a_{m+1}=(1+a_{m+2})^{m+1}\geqslant 0$$ which is impossible (the last inequality would follow from $m+1$ being even). This establishes that for all $n$ we must have $a_n>-1$. From Bernoulli inequality then we obtain $$(1+a_{n+1})^n=na_n+1\leqslant(a_n+1)^n\Rightarrow a_{n+1}\leqslant a_n$$ Hence a monotonic decreasing sequence of real numbers. Since $a_n\leqslant a_2$ and $a_n>-1$ for all $n$ the sequence is also bounded. Therefore by Bolzano-Weierstrass theorem (bounded and monotone sequences converge) then $\lim_na_n=a$ for some $a<+\infty$. Next we show that $a=0$. Suppose first that $a>0$ then $$(1+a)^n\leqslant(1+a_{n+1})^n=na_n+1\leqslant na_2+1$$ but this cannot hold for arbitrarily large $n$. The left side grows exponentially while the right linearly. Now assume that $a<0$ then there is some $N\in\mathbb{N}$ such that for all $n\geqslant N$ we have $a_n<0$ but then $$\lim_n(1+a_{n+1})^n=0\Rightarrow \lim_n(na_n)=-1\Rightarrow \lim_na_n=0$$ which raises a contradiction. Therefore indeed $\lim_na_n=a=0$. This gives also the additional information that the sequence $(a_n)$ is a nonnegative sequence (since it is monotonic decreasing). Next we show $\lim_na_{n+1}/a_n=1$. $$1\geqslant\lim_n\frac{a_{n+1}}{a_n}=\lim_n\frac{\sqrt[n]{na_n+1}-1}{a_n}=\lim_n\frac{(na_n+1)-1}{a_n((na_n+1)^{(n-1)/n}+...+(na_n+1)^{2/n}+(na_n+1)^{1/n}+1)}\\=\lim_n\frac{n}{((na_n+1)^{(n-1)/n}+...+(na_n+1)^{2/n}+(na_n+1)^{1/n}+1)}\\ \geqslant \lim_n\frac{n}{n(na_n+1)^{(n-1)/n}}=\lim_n\frac{a_{n+1}+1}{na_n+1}=\frac{\lim_na_{n+1}+1}{\lim_nna_n+1}=\frac{1}{\lim_nna_n+1}=1$$ whenever $\lim_nna_n=0$. On the other hand we have $$\lim_ne^{a_{n+1}n}=\lim_n(1+a_{n+1})^n=\lim_n(na_n)+1\leqslant\lim_n(1+a_n)^n=\lim_ne^{a_nn}$$ First if $\lim_nna_n=+\infty$ then the above relation cannot hold as the side terms grow exponentially and the middle term $na_n+1$ grows linearly. So let $\lim_nna_n=L<+\infty$ then if $\lim_na_{n}/a_{n+1}=1$ we obtain $$L=\lim_na_nn=\lim_n\frac{a_{n}}{a_{n+1}}a_{n+1}n=\lim_n\frac{a_n}{a_{n+1}}\lim_na_{n+1}n=\lim_na_{n+1}n$$ this implies $$e^L=\lim_ne^{a_{n+1}n}=\lim_nna_n+1=L+1\leqslant\lim_ne^{na_n}= e^L\Rightarrow L=0$$

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