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# ?

Question

Cameron Alexander · Accepted Answer

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

Assuming we want to deal with Real numbers only, we require #y >= 0# If #y=0# then the radius of convergence is infinite. Otherwise, #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#.

Axel Alvarez · Answer

undefined To determine the convergence radius of the series $ \sum_{k=1}^{\infty} a_k x^k $, where $ a_k = \sqrt{y + \sqrt{y + \sqrt{y + \ldots}}} $ and $ a_1 = \sqrt{y} $, we first need to find a recursive formula for $ a_k $. This will help us express each term $ a_k $ in terms of the preceding term $ a_{k-1} $.

Given the sequence, we can observe that $ a_k = \sqrt{y + a_{k-1}} $.

To find the convergence radius of the series, we'll use the Ratio Test. The Ratio Test states that if $ L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| $, then the series converges absolutely if $ L < 1 $, diverges if $ L > 1 $, and the test is inconclusive if $ L = 1 $.

Now, let's compute the ratio $ \frac{a_{k+1}}{a_k} $:

$$
\frac{a_{k+1}}{a_k} = \frac{\sqrt{y + a_k}}{a_k}
$$

Using the recursive formula for $ a_k $, we substitute $ a_k = \sqrt{y + a_{k-1}} $:

$$
= \frac{\sqrt{y + \sqrt{y + a_{k-1}}}}{\sqrt{y + a_{k-1}}}
$$

$$
= \frac{\sqrt{y + \sqrt{y + a_{k-1}}}}{\sqrt{y + a_{k-1}}} \cdot \frac{\sqrt{y + a_{k-1}}}{\sqrt{y + a_{k-1}}}
$$

$$
= \frac{\sqrt{y + \sqrt{y + a_{k-1}}}}{y + a_{k-1}}
$$

Now, let's take the limit as $ k $ approaches infinity:

$$
L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|
$$

$$
= \lim_{k \to \infty} \left| \frac{\sqrt{y + \sqrt{y + a_{k-1}}}}{y + a_{k-1}} \right|
$$

$$
= \lim_{k \to \infty} \left| \frac{\sqrt{y + \sqrt{y + \ldots}}}{y + \sqrt{y + \ldots}} \right|
$$

Since this limit converges to 1, the Ratio Test is inconclusive.

Thus, we need to use another method to determine the convergence radius. We can use the Cauchy-Hadamard theorem, which states that the radius of convergence $ R $ is given by:

$$
R = \frac{1}{\limsup_{k \to \infty} \sqrt[k]{|a_k|}}
$$

We can find $ \limsup_{k \to \infty} \sqrt[k]{|a_k|} $ by considering the sequence $ \sqrt{y}, \sqrt{y + \sqrt{y}}, \sqrt{y + \sqrt{y + \sqrt{y}}}, \ldots $.

This sequence is increasing and bounded above by $ y + 1 $. Thus, $ \limsup_{k \to \infty} \sqrt[k]{|a_k|} = y + 1 $.

Therefore, the convergence radius $ R $ is:

$$
R = \frac{1}{y + 1}
$$