Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
- Differential equations Step by Step
- Product of series
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How to use it?
- Sum of series:
- (n+2/2n)^n2
- ln(5k/n)
- factorial(n)*n^(-p)*factorial(q-1)/factorial(q+n)
-
(-1)^nsin(1/n)
-
Identical expressions
- ((- one)^(k+ one)*(two ^(two k- one))*x^(two *k))/factorial(2*k)
- (( minus 1) to the power of (k plus 1) multiply by (2 to the power of (2k minus 1)) multiply by x to the power of (2 multiply by k)) divide by factorial(2 multiply by k)
- (( minus one) to the power of (k plus one) multiply by (two to the power of (two k minus one)) multiply by x to the power of (two multiply by k)) divide by factorial(2 multiply by k)
- ((-1)(k+1)*(2(2k-1))*x(2*k))/factorial(2*k)
- -1k+1*22k-1*x2*k/factorial2*k
- ((-1)^(k+1)(2^(2k-1))x^(2k))/factorial(2k)
- ((-1)(k+1)(2(2k-1))x(2k))/factorial(2k)
- -1k+122k-1x2k/factorial2k
- -1^k+12^2k-1x^2k/factorial2k
- ((-1)^(k+1)*(2^(2k-1))*x^(2*k)) divide by factorial(2*k)
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Similar expressions
- ((-1)^(k-1)*(2^(2k-1))*x^(2*k))/factorial(2*k)
- ((1)^(k+1)*(2^(2k-1))*x^(2*k))/factorial(2*k)
- ((-1)^(k+1)*(2^(2k+1))*x^(2*k))/factorial(2*k)
Sum of series ((-1)^(k+1)*(2^(2k-1))*x^(2*k))/factorial(2*k)
The solution
You have entered
[src]
oo ____ \ ` \ k + 1 2*k - 1 2*k \ (-1) *2 *x / ----------------------- / (2*k)! /___, k = 1
$$\sum_{k=1}^{\infty} \frac{x^{2 k} \left(-1\right)^{k + 1} \cdot 2^{2 k - 1}}{\left(2 k\right)!}$$
Sum((((-1)^(k + 1)*2^(2*k - 1))*x^(2*k))/factorial(2*k), (k, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{x^{2 k} \left(-1\right)^{k + 1} \cdot 2^{2 k - 1}}{\left(2 k\right)!}$$
It is a series of species
$$a_{k} \left(c x - x_{0}\right)^{d k}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{k \to \infty} \left|{\frac{a_{k}}{a_{k + 1}}}\right|}{c}$$
In this case
$$a_{k} = \frac{\left(-1\right)^{k + 1} \cdot 2^{2 k - 1}}{\left(2 k\right)!}$$
and
$$x_{0} = 0$$
,
$$d = 2$$
,
$$c = 1$$
then
$$R^{2} = \lim_{k \to \infty}\left(2^{- 2 k - 1} \cdot 2^{2 k - 1} \left|{\frac{\left(2 k + 2\right)!}{\left(2 k\right)!}}\right|\right)$$
Let's take the limit
we find
$$R^{2} = \infty$$
$$R = \infty$$
$$\frac{x^{2 k} \left(-1\right)^{k + 1} \cdot 2^{2 k - 1}}{\left(2 k\right)!}$$
It is a series of species
$$a_{k} \left(c x - x_{0}\right)^{d k}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{k \to \infty} \left|{\frac{a_{k}}{a_{k + 1}}}\right|}{c}$$
In this case
$$a_{k} = \frac{\left(-1\right)^{k + 1} \cdot 2^{2 k - 1}}{\left(2 k\right)!}$$
and
$$x_{0} = 0$$
,
$$d = 2$$
,
$$c = 1$$
then
$$R^{2} = \lim_{k \to \infty}\left(2^{- 2 k - 1} \cdot 2^{2 k - 1} \left|{\frac{\left(2 k + 2\right)!}{\left(2 k\right)!}}\right|\right)$$
Let's take the limit
we find
$$R^{2} = \infty$$
$$R = \infty$$
The answer
[src]
2 / 1 cos(2*x)\ x *|---- - --------| | 2 2 | \2*x 2*x /
$$x^{2} \left(- \frac{\cos{\left(2 x \right)}}{2 x^{2}} + \frac{1}{2 x^{2}}\right)$$
x^2*(1/(2*x^2) - cos(2*x)/(2*x^2))
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