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Consider the sequence $\{a_{n}\}$, with $n\ge1$ and $a>0$, defined as:

$$a_{n}=\sqrt{1+\sqrt{a+\sqrt{a^2+\cdots+\sqrt{a^n}}}}$$

I'm trying to prove here 2 things: a). the sequence is convergent; b). the sequence's limit when n goes to $\infty$. I may suppose that there must be a proof for this general case. I saw this problem with the case $a=2$ (where it was required to prove only the convergence), but this is just a particular case. The generalization seems to be much more interesting.

Michael Hardy
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user 1591719
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    In the special case $a = 4$, the limit of the sequence is $2$. This is because $2^x + 1 = \sqrt{4^x + \sqrt{4^{x+1} + \sqrt{4^{x+2} + \cdots }}}$ whenever the radical converges, which in turn can be verified heuristically with the functional equation $f(x)^2 = 4^x + f(x + 1)$ satisfied by $f(x) = 2^x + 1$. However, the argument is specific to $a = 4$. – Nick Strehlke Jul 14 '12 at 23:12
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    Related [Nested radical $\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$](https://math.stackexchange.com/questions/1683762/nested-radical-sqrtx-sqrtx2-cdots-sqrtxn-cdots) – Sil Jul 06 '18 at 10:46

2 Answers2

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Here is a full answer to part (a) and a partial answer to part (b). Call $(a_n(a))_{n\geqslant1}$ the sequence when the value of the parameter is $a$.

One has $a_0(1)=1$ and $a_{n+1}(1)=u(a_n(1))$ for every $n\geqslant0$ with $u(x)=\sqrt{1+x}$. Hence the usual technique shows that the sequence $(a_n(1))_{n\geqslant0}$ is increasing to $a_\infty(1)=\alpha$ where $\alpha$ solves the equation $\alpha=u(\alpha)$, that is, $\color{red}{\alpha=\frac12(1+\sqrt5)}$.

When $a\lt1$, $a_n(a)\leqslant a_n(1)$ and $(a_n(a))_{n\geqslant0}$ is increasing hence $(a_n(a))_{n\geqslant0}$ converges to a finite limit $a_\infty(a)$ with $\color{red}{\sqrt{1+\sqrt{a}}\lt a_\infty(a)\leqslant \alpha}$.

When $a\gt1$, $\sqrt{1+\sqrt{aa_{n-1}(1)}}\leqslant a_n(a)\leqslant\sqrt{1+\sqrt{a}a_{n-1}(1)}$ and $(a_n(a))_{n\geqslant0}$ is increasing hence it converges to a finite limit $a_\infty(a)$ with $\color{red}{\sqrt{1+\sqrt{\alpha a}}\lt a_\infty(a)\leqslant\sqrt{1+\alpha\sqrt{a}}}$.

To show the upper bound on $a_n(a)$, one carries over every power of $a$ to the left until it reaches the position of $\sqrt{a}$. Crossing a square root sign halves the exponent and $a\gt1$ hence the power of $a$ which just crossed a square root sign is smaller than the preceding one. For example, the first step of the proof uses $a^{n/2}\leqslant a^{n-1}$ to deduce $$ \sqrt{a^{n-1}+\sqrt{a^n}}=\sqrt{a^{n-1}+a^{n/2}\sqrt1}\leqslant\sqrt{a^{n-1}(1+\sqrt1)}=a^{(n-1)/2}\sqrt{1+\sqrt1}, $$ the second step uses $a^{(n-1)/2}\leqslant a^{n-2}$, and so on, until $a^{3/2}\leqslant a^2$ and $a^{2/2}\leqslant a$. A similar reasoning yields the lower bound.

Finally, the map $a\mapsto a_\infty(a)$ is nondecreasing from $\color{red}{a_\infty(0)=1}$ to $\color{red}{a_\infty(+\infty)=+\infty}$.

Did
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The Herschfeld's Convergence Theorem may help (the more general subject is 'Nested Radical').

Herschfeld's paper "On Infinite Radicals" is available here.

Raymond Manzoni
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