Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
- Differential equations Step by Step
- Product of series
-
How to use it?
- Sum of series:
- x^n/n^2
-
7+k
-
log(1+1/n)
- x^(n+1)/(n(n+1))
-
Identical expressions
- (- one)^n*pi^(2n+ one)/((2n+ one)*(2n+ one)!)
- ( minus 1) to the power of n multiply by Pi to the power of (2n plus 1) divide by ((2n plus 1) multiply by (2n plus 1)!)
- ( minus one) to the power of n multiply by Pi to the power of (2n plus one) divide by ((2n plus one) multiply by (2n plus one)!)
- (-1)n*pi(2n+1)/((2n+1)*(2n+1)!)
- -1n*pi2n+1/2n+1*2n+1!
- (-1)^npi^(2n+1)/((2n+1)(2n+1)!)
- (-1)npi(2n+1)/((2n+1)(2n+1)!)
- -1npi2n+1/2n+12n+1!
- -1^npi^2n+1/2n+12n+1!
- (-1)^n*pi^(2n+1) divide by ((2n+1)*(2n+1)!)
-
Similar expressions
- (1)^n*pi^(2n+1)/((2n+1)*(2n+1)!)
- (-1)^n*pi^(2n+1)/((2n+1)*(2n-1)!)
- (-1)^n*pi^(2n-1)/((2n+1)*(2n+1)!)
- (-1)^n*pi^(2n+1)/((2n-1)*(2n+1)!)
Sum of series (-1)^n*pi^(2n+1)/((2n+1)*(2n+1)!)
The solution
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oo ____ \ ` \ n 2*n + 1 \ (-1) *pi / -------------------- / (2*n + 1)*(2*n + 1)! /___, n = 0
$$\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n} \pi^{2 n + 1}}{\left(2 n + 1\right) \left(2 n + 1\right)!}$$
Sum(((-1)^n*pi^(2*n + 1))/(((2*n + 1)*factorial(2*n + 1))), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\frac{\left(-1\right)^{n} \pi^{2 n + 1}}{\left(2 n + 1\right) \left(2 n + 1\right)!}$$
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
$$a_{n} = \frac{\pi^{2 n + 1}}{\left(2 n + 1\right) \left(2 n + 1\right)!}$$
and
$$x_{0} = 1$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(1 + \lim_{n \to \infty}\left(\frac{\pi^{- 2 n - 3} \pi^{2 n + 1} \left(2 n + 3\right) \left|{\frac{\left(2 n + 3\right)!}{\left(2 n + 1\right)!}}\right|}{2 n + 1}\right)\right)$$
Let's take the limit
we find
$$R = \infty$$
$$\frac{\left(-1\right)^{n} \pi^{2 n + 1}}{\left(2 n + 1\right) \left(2 n + 1\right)!}$$
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
$$a_{n} = \frac{\pi^{2 n + 1}}{\left(2 n + 1\right) \left(2 n + 1\right)!}$$
and
$$x_{0} = 1$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(1 + \lim_{n \to \infty}\left(\frac{\pi^{- 2 n - 3} \pi^{2 n + 1} \left(2 n + 3\right) \left|{\frac{\left(2 n + 3\right)!}{\left(2 n + 1\right)!}}\right|}{2 n + 1}\right)\right)$$
Let's take the limit
we find
$$R = \infty$$
The rate of convergence of the power series
The graph
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