A Limit of Another Sequence

Leonard Giugiuc, Dan Sitaru
15 December, 2015

Let $\{a_n\},$ $n\ge 1,$ with $a_1\gt 0$ and $a_{n+1}=a_n(4+a_n),$ for $n\ge 1.$ Find

$\displaystyle \lim_{n\rightarrow\infty}\sqrt[2^{n}]{a_{n}}.$

The problem is due to Dan Nedeianu and was published in Gazeta Matematica.

Solution

Obviously, $a_n\gt 0,$ for $n\ge 1.$ We have, $\displaystyle\sqrt{a_{n+1}}=\sqrt{a_n}\sqrt{4+a_n}.$ If $x_n=\sqrt{a_n}$ then $x_{n+1}=x_n\cdot\sqrt{4+x^2_n}$ and $x_1\gt 0.$

Introduce $\displaystyle t=\ln\left(\frac{\sqrt{4+x_1^2}+x_1}{2}\right)$ so that $x_1=2\sinh t.$

We observe $x_2=2\sinh 2t$ and, by an easy induction, $x_n=2\sinh 2^{n-1}t.$ From here, $\displaystyle\sqrt[\displaystyle 2^n]{a_n}=2^{^{\displaystyle{\frac{1}{2^{n-1}}}}}\left(\sinh 2^{n-1}t\right)^{^{\displaystyle{\frac{1}{2^{n-1}}}}}.$ But $\displaystyle\lim_{n\rightarrow\infty}2^{^{\displaystyle{\frac{1}{2^{n-1}}}}}=1.$ Thus, suffice it to find $\displaystyle\lim_{n\rightarrow\infty}\frac{\ln\left(\sinh 2^{n-1}t\right)}{2^{n-1}}.$ By l'Hôpital's rule, for $x=2^{n-1},$ we obtain that that limit equals $t:$ $\displaystyle\lim_{n\rightarrow\infty}\frac{\ln\left(\sinh 2^{n-1}t\right)}{2^{n-1}}=t,$ implying

$\displaystyle\lim_{n\rightarrow\infty}\sqrt[\displaystyle 2^n]{a_{n}}=e^t=\frac{\sqrt{4+x_1^2}+x_1}{2}=\frac{\sqrt{4+a_1}+\sqrt{a_1}}{2}.$

(This was posted at the CutTheKnotMath facebook page.)

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