# Find the​ series' interval of convergence​ and, within this​ interval, the sum of the series as a function of x. sum_{n=0}^inftyfrac{(x-4)^{2n}}{36^n}

Question
Series
Find the​ series' interval of convergence​ and, within this​ interval, the sum of the series as a function of x.
$$\sum_{n=0}^\infty\frac{(x-4)^{2n}}{36^n}$$

2021-01-06
Given series is
$$\sum_{n=0}^\infty\frac{(x-4)^{2n}}{36^n}$$
Now,
let $$(x-4)^2=y$$
So, the series becomes
$$\sum_{n=0}^\infty\frac{y^n}{36^n}$$
So, $$a_n=\frac{1}{36^n}$$
Then,
$$(a_n)^\frac{1}{n}=\frac{1}{36}$$
and
$$\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\frac{1}{36}$$
So, radius of convergence is 36
Then, the radius of convergence of the given series is 6
Interval of convergence $$=(-6-4,-6+4)=(-10,-2)$$
Now,
$$a_0=1$$
$$a_1=\frac{y}{36}$$
$$a_2=\frac{y^2}{36^2}$$
$$\vdots$$
$$a_n=\frac{y^n}{36^n}$$
So, $$S_n=a_0+a_1+...+a_n$$
$$=1+\frac{y}{36}+\frac{y^2}{36^2}+...+\frac{y^n}{36^n}$$
$$=\frac{1[1-(\frac{y}{36})^n]}{1-\frac{y}{36}}$$
$$=\frac{36[1-\frac{(x-4)^{2n}}{36^n}]}{36-(x-4)^2}$$
$$=\frac{36^n-(x-4)^{2n}}{36^{n-1}[36-(x-4)^2]}$$
Hence, the sum of the series as a function of x is
$$=\frac{36^n-(x-4)^{2n}}{36^{n-1}[36-(x-4)^2]}$$

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