Given the sequence $a_{1} = \sqrt{y}, a_{2} = \sqrt{y + \sqrt{y}}, a_{3} = \sqrt{y + \sqrt{y + \sqrt{y}}}, \dots$ $a_1=sqrt(y),a_2=sqrt(y+sqrt(y)), a_3 = sqrt(y+sqrt(y+sqrt(y))), cdots$ determine the convergence radius of $\infty \sum k = 1 a_{k} x^{k}$ $sum_(k=1)^oo a_k x^k$ ?

Question

1698 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-08T20:39:50

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

Assuming we want to deal with Real numbers only, we require $y \geq 0$ $y >= 0$

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

Otherwise, $y > 0$ $y > 0$

Note that the sequence $a_1, a_2, a_3,...$ is strictly monotonic increasing.

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

Let $t = sqrt(y+sqrt(y+sqrt(y+sqrt(y+sqrt(y+...)))))$

Then:

$t^2-t-y = 0$

So using the quadratic formula:

$t = (1+-sqrt(1+4y))/2$

and since $t >= 0$ we must have:

$t = 1/2+sqrt(1+4y)/2 = 1/2+sqrt(y+1/4)$

This is a fixed point of the function $f(t) = sqrt(y+t)$

In particular, if $x >= 0$ then

$sqrt(y) sum_(k=1)^oo x^k <= sum_(k=1)^oo a_k x^k <= (1/2+sqrt(y+1/4)) sum_(k=1)^oo x^k$

So $sum_(k=1)^oo a_k x^k$ is absolutely convergent if and only if $sum_(k=1)^oo x^k$ is absolutely convergent, which is if and only if $abs(x) < 1$ .

Hence the radius of convergence is $1$ .

Given the sequence a_1=sqrt(y),a_2=sqrt(y+sqrt(y)), a_3 = sqrt(y+sqrt(y+sqrt(y))), cdotsa1=√y,a2=√y+√y,a3=√y+√y+√y,⋯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∞∑k=1akxksum_(k=1)^oo a_k x^k ?