Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
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- Product of series
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How to use it?
- Sum of series:
- tg^(2)*(6/(2n)^(1/3))*(x-24)^n
-
1/(2*n-1)*(2*n+1)
- cos(kx)
- ((-1)^(n-1))*(5^n+2)*x^n
-
Identical expressions
- cos* two *n*x/ four *n^ two - one
- co sinus of e of multiply by 2 multiply by n multiply by x divide by 4 multiply by n squared minus 1
- co sinus of e of multiply by two multiply by n multiply by x divide by four multiply by n to the power of two minus one
- cos*2*n*x/4*n2-1
- cos*2*n*x/4*n²-1
- cos*2*n*x/4*n to the power of 2-1
- cos2nx/4n^2-1
- cos2nx/4n2-1
- cos*2*n*x divide by 4*n^2-1
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Similar expressions
- cos*2*n*x/4*n^2+1
Sum of series cos*2*n*x/4*n^2-1
The solution
You have entered
[src]
oo ___ \ ` \ /cos(2*n)*x 2 \ ) |----------*n - 1| / \ 4 / /__, n = 1
$$\sum_{n=1}^{\infty} \left(n^{2} \frac{x \cos{\left(2 n \right)}}{4} - 1\right)$$
Sum(((cos(2*n)*x)/4)*n^2 - 1, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$n^{2} \frac{x \cos{\left(2 n \right)}}{4} - 1$$
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
$$a_{n} = \frac{n^{2} x \cos{\left(2 n \right)}}{4} - 1$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\frac{n^{2} x \cos{\left(2 n \right)}}{4} - 1}{\frac{x \left(n + 1\right)^{2} \cos{\left(2 n + 2 \right)}}{4} - 1}}\right|$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty} \left|{\frac{\frac{n^{2} x \cos{\left(2 n \right)}}{4} - 1}{\frac{x \left(n + 1\right)^{2} \cos{\left(2 n + 2 \right)}}{4} - 1}}\right|$$
$$n^{2} \frac{x \cos{\left(2 n \right)}}{4} - 1$$
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
$$a_{n} = \frac{n^{2} x \cos{\left(2 n \right)}}{4} - 1$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\frac{n^{2} x \cos{\left(2 n \right)}}{4} - 1}{\frac{x \left(n + 1\right)^{2} \cos{\left(2 n + 2 \right)}}{4} - 1}}\right|$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty} \left|{\frac{\frac{n^{2} x \cos{\left(2 n \right)}}{4} - 1}{\frac{x \left(n + 1\right)^{2} \cos{\left(2 n + 2 \right)}}{4} - 1}}\right|$$
False
The answer
[src]
oo ____ \ ` \ / 2 \ \ | x*n *cos(2*n)| / |-1 + -------------| / \ 4 / /___, n = 1
$$\sum_{n=1}^{\infty} \left(\frac{n^{2} x \cos{\left(2 n \right)}}{4} - 1\right)$$
Sum(-1 + x*n^2*cos(2*n)/4, (n, 1, oo))
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