Given the sequence a_1=sqrt(y),a_2=sqrt(y+sqrt(y)), a_3 = sqrt(y+sqrt(y+sqrt(y))), cdots determine the convergence radius of sum_(k=1)^oo a_k x^k ?

Sep 8, 2016

If $y = 0$ then the radius is infinite. Otherwise it is $1$.

Explanation:

Assuming we want to deal with Real numbers only, we require $y \ge 0$

If $y = 0$ then the radius of convergence is infinite.

Otherwise, $y > 0$

Note that the sequence ${a}_{1} , {a}_{2} , {a}_{3} , \ldots$ is strictly monotonic increasing.

It does have a finite fixed point towards which it converges:

Let $t = \sqrt{y + \sqrt{y + \sqrt{y + \sqrt{y + \sqrt{y + \ldots}}}}}$

Then:

${t}^{2} - t - y = 0$

$t = \frac{1 \pm \sqrt{1 + 4 y}}{2}$

and since $t \ge 0$ we must have:

$t = \frac{1}{2} + \frac{\sqrt{1 + 4 y}}{2} = \frac{1}{2} + \sqrt{y + \frac{1}{4}}$

This is a fixed point of the function $f \left(t\right) = \sqrt{y + t}$

In particular, if $x \ge 0$ then

$\sqrt{y} {\sum}_{k = 1}^{\infty} {x}^{k} \le {\sum}_{k = 1}^{\infty} {a}_{k} {x}^{k} \le \left(\frac{1}{2} + \sqrt{y + \frac{1}{4}}\right) {\sum}_{k = 1}^{\infty} {x}^{k}$

So ${\sum}_{k = 1}^{\infty} {a}_{k} {x}^{k}$ is absolutely convergent if and only if ${\sum}_{k = 1}^{\infty} {x}^{k}$ is absolutely convergent, which is if and only if $\left\mid x \right\mid < 1$.

Hence the radius of convergence is $1$.