A Limit of a Sequence

Leonard Giugiuc, Dan Sitaru
11 December, 2015

Solution

Obviously, $a_n\gt 0,$ for $n\ge 1.$ Also, $\displaystyle\frac{a_{n+1}}{a_n}=4+a_n\gt 1,$ to say the least. Thus, $\{a_n\}$ is strictly increasing to infinity.

Introduce $x_n=a_n+2$ and $b_n=\sqrt[2^n]{a_n},$ $n\ge 1.$

We have $x_1\gt 2$ and, for $n\ge 1,$ $\displaystyle x_{n+1}=x^{2}_{n}-2;$ from here, $x_{n+1}\lt x^{2}_{n},$ and, consequently, for $n\ge 2,$ $\displaystyle x_n\lt x_{1}^{2^{n-1}},$ implying $\displaystyle x_n-2 \lt x_{1}^{2^{n-1}},$ such that $a_n\lt (2+a_1)^{2^{n-1}}.$ It follows that $\{b_n\}$ is bounded from above: $b_n\lt\sqrt{2+a_1},$ $n\ge 2.$

On the other hand, $\displaystyle\frac{a_{n+1}}{a_n^2}=\frac{4+a_n}{a_n}\gt 1,$ i.e., $\displaystyle\left(\frac{b_{n+1}}{b_n}\right)^{2^{n+1}}\gt 1$ and, therefore, $\displaystyle\frac{b_{n+1}}{b_n}\gt 1,$ making $\{b_n\}$ monotone increasing. It follows that $\{b_n\}$ has a limit, say $\displaystyle b=\lim_{n\rightarrow\infty}b_n.$

Further, $\displaystyle 1=\lim_{n\rightarrow\infty}\frac{4+a_n}{a_n}=\lim_{n\rightarrow\infty}\left(\frac{b_{n+1}}{b_n}\right)^{2^{n+1}}=e^{\displaystyle\lim_{n\rightarrow\infty}\frac{2^{n+1}(b_{n+1}-b_{n})}{b_n}}.$ Thus $\displaystyle\lim_{n\rightarrow\infty}\frac{2^{n+1}(b_{n+1}-b_{n})}{b_{n}}=0,$ and since $\displaystyle\lim_{n\rightarrow\infty}\frac{2}{b_n}=\frac{2}{b},$ we get the required $\displaystyle \lim_{n\rightarrow\infty}\left[2^n\left(\sqrt[2^{n+1}]{a_{n+1}}-\sqrt[2^n]{a_n}\right)\right]=0.$

(This was posted at the CutTheKnotMath facebook page. My only question is about the role played by $4$ in the definition. Dan and Leo kindly posted another limit which they modified to develop the present one. This was the original appearance of $4.)$

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