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Sum of series (4x)^(2n)
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oo ___ \ ` \ 2*n / (4*x) /__, n = 0
$$\sum_{n=0}^{\infty} \left(4 x\right)^{2 n}$$
Sum((4*x)^(2*n), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\left(4 x\right)^{2 n}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = 1$$
and
$$x_{0} = 0$$
,
$$d = 2$$
,
$$c = 4$$
then
$$R^{2} = \frac{\lim_{n \to \infty} 1}{4}$$
Let's take the limit
we find
$$R^{2} = \frac{1}{4}$$
$$R = 0.5$$
$$\left(4 x\right)^{2 n}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = 1$$
and
$$x_{0} = 0$$
,
$$d = 2$$
,
$$c = 4$$
then
$$R^{2} = \frac{\lim_{n \to \infty} 1}{4}$$
Let's take the limit
we find
$$R^{2} = \frac{1}{4}$$
$$R = 0.5$$
The answer
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/ 1 | 2| | --------- for 16*|x | < 1 | 2 | 1 - 16*x | | oo < ___ | \ ` | \ 2*n 2*n | / 4 *x otherwise | /__, |n = 0 \
$$\begin{cases} \frac{1}{1 - 16 x^{2}} & \text{for}\: 16 \left|{x^{2}}\right| < 1 \\\sum_{n=0}^{\infty} 4^{2 n} x^{2 n} & \text{otherwise} \end{cases}$$
Piecewise((1/(1 - 16*x^2), 16*|x^2| < 1), (Sum(4^(2*n)*x^(2*n), (n, 0, oo)), True))
Examples of finding the sum of a series
- The Sum of the Power Series
x^n/n
(x-1)^n
- Factorial
1/2^(n!)
n^2/n!
x^n/n!
k!/(n!*(n+k)!)
- Flint Hills Series
csc(n)^2/n^3
- Basel problem
1/n^2
1/n^4
1/n^6
- Harmonic series
1/n
- Grandi's series
(-1)^n
- Alternating series
(-1)^(n + 1)/n
(n + 2)*(-1)^(n - 1)
(3*n - 1)/(-5)^n
(-1)^(n - 1)*n/(6*n - 5)
- Newton–Mercator series
(-1)^(n + 1)/n*x^n
- Exam the series for convergence
(3*n - 1)/(-5)^n