Mister Exam
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How to use it?
- Sum of series:
-
sqrt((n+4)/((n^4)+4))
-
(7^n-3^n)/21^n
-
sin(pi/2^(n-1))
- sin(n*x)/n^2
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Identical expressions
- (four n)!/(n!)^ four *(x/4)^(n+ one)
- (4n)! divide by (n!) to the power of 4 multiply by (x divide by 4) to the power of (n plus 1)
- (four n)! divide by (n!) to the power of four multiply by (x divide by 4) to the power of (n plus one)
- (4n)!/(n!)4*(x/4)(n+1)
- 4n!/n!4*x/4n+1
- (4n)!/(n!)⁴*(x/4)^(n+1)
- (4n)!/(n!)^4(x/4)^(n+1)
- (4n)!/(n!)4(x/4)(n+1)
- 4n!/n!4x/4n+1
- 4n!/n!^4x/4^n+1
- (4n)! divide by (n!)^4*(x divide by 4)^(n+1)
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Similar expressions
- (4n)!/(n!)^4*(x/4)^(n-1)
Sum of series (4n)!/(n!)^4*(x/4)^(n+1)
The solution
You have entered
[src]
oo ____ \ ` \ n + 1 \ (4*n)! /x\ ) ------*|-| / 4 \4/ / n! /___, n = 0
$$\sum_{n=0}^{\infty} \left(\frac{x}{4}\right)^{n + 1} \frac{\left(4 n\right)!}{n!^{4}}$$
Sum((factorial(4*n)/factorial(n)^4)*(x/4)^(n + 1), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\left(\frac{x}{4}\right)^{n + 1} \frac{\left(4 n\right)!}{n!^{4}}$$
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
$$a_{n} = \frac{\left(\frac{x}{4}\right)^{n + 1} \left(4 n\right)!}{n!^{4}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left|{\frac{\left(\frac{x}{4}\right)^{- n - 2} \left(\frac{x}{4}\right)^{n + 1} \left(4 n\right)!}{n!^{4} \left(4 n + 4\right)!}}\right| \left(n + 1\right)!^{4}\right)$$
Let's take the limit
we find
$$1 = \frac{1}{64 \left|{x}\right|}$$
$$1 = \frac{0.015625}{\left|{x}\right|}$$
$$\left(\frac{x}{4}\right)^{n + 1} \frac{\left(4 n\right)!}{n!^{4}}$$
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
$$a_{n} = \frac{\left(\frac{x}{4}\right)^{n + 1} \left(4 n\right)!}{n!^{4}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left|{\frac{\left(\frac{x}{4}\right)^{- n - 2} \left(\frac{x}{4}\right)^{n + 1} \left(4 n\right)!}{n!^{4} \left(4 n + 4\right)!}}\right| \left(n + 1\right)!^{4}\right)$$
Let's take the limit
we find
$$1 = \frac{1}{64 \left|{x}\right|}$$
$$1 = \frac{0.015625}{\left|{x}\right|}$$
False
The answer
[src]
// _ \
|| |_ /1/4, 1/2, 3/4 | \ |
|| | | | 64*x| for 64*|x| <= 1|
||3 2 \ 1, 1 | / |
|| |
|| oo |
|| ____ |
x*|< \ ` |
|| \ -n n |
|| \ 4 *x *(4*n)! |
|| ) ------------- otherwise |
|| / 4 |
|| / n! |
|| /___, |
\\ n = 0 /
-------------------------------------------------
4
$$\frac{x \left(\begin{cases} {{}_{3}F_{2}\left(\begin{matrix} \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ 1, 1 \end{matrix}\middle| {64 x} \right)} & \text{for}\: 64 \left|{x}\right| \leq 1 \\\sum_{n=0}^{\infty} \frac{4^{- n} x^{n} \left(4 n\right)!}{n!^{4}} & \text{otherwise} \end{cases}\right)}{4}$$
x*Piecewise((hyper((1/4, 1/2, 3/4), (1, 1), 64*x), 64*|x| <= 1), (Sum(4^(-n)*x^n*factorial(4*n)/factorial(n)^4, (n, 0, oo)), True))/4
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