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:
-
(4*9^n-2^n)/18^n
-
1/((3*n+1)*(3*n+4))
- sqrt(n)*(x+3)^n/n^2+1
-
ln((n(n+2))/(n+1)^2)
-
Identical expressions
- ((- one)**(n- one))*((five **(two *n- one))/(two *n- one))
- (( minus 1) multiply by multiply by (n minus 1)) multiply by ((5 multiply by multiply by (2 multiply by n minus 1)) divide by (2 multiply by n minus 1))
- (( minus one) multiply by multiply by (n minus one)) multiply by ((five multiply by multiply by (two multiply by n minus one)) divide by (two multiply by n minus one))
- ((-1)(n-1))((5(2n-1))/(2n-1))
- -1n-152n-1/2n-1
- ((-1)**(n-1))*((5**(2*n-1)) divide by (2*n-1))
-
Similar expressions
- ((-1)**(n-1))*((5**(2*n+1))/(2*n-1))
- ((-1)**(n-1))*((5**(2*n-1))/(2*n+1))
- ((-1)**(n+1))*((5**(2*n-1))/(2*n-1))
- ((1)**(n-1))*((5**(2*n-1))/(2*n-1))
Sum of series ((-1)**(n-1))*((5**(2*n-1))/(2*n-1))
The solution
You have entered
[src]
oo ____ \ ` \ 2*n - 1 \ n - 1 5 / (-1) *-------- / 2*n - 1 /___, n = 0
$$\sum_{n=0}^{\infty} \left(-1\right)^{n - 1} \frac{5^{2 n - 1}}{2 n - 1}$$
Sum((-1)^(n - 1)*(5^(2*n - 1)/(2*n - 1)), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\left(-1\right)^{n - 1} \frac{5^{2 n - 1}}{2 n - 1}$$
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
$$a_{n} = \frac{\left(-1\right)^{n - 1} \cdot 5^{2 n - 1}}{2 n - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(5^{- 2 n - 1} \cdot 5^{2 n - 1} \left(2 n + 1\right) \left|{\frac{1}{2 n - 1}}\right|\right)$$
Let's take the limit
we find
$$\left(-1\right)^{n - 1} \frac{5^{2 n - 1}}{2 n - 1}$$
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
$$a_{n} = \frac{\left(-1\right)^{n - 1} \cdot 5^{2 n - 1}}{2 n - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(5^{- 2 n - 1} \cdot 5^{2 n - 1} \left(2 n + 1\right) \left|{\frac{1}{2 n - 1}}\right|\right)$$
Let's take the limit
we find
False
False
False
The rate of convergence of the power series
The answer
[src]
oo ____ \ ` \ -1 + n -1 + 2*n \ (-1) *5 / -------------------- / -1 + 2*n /___, n = 0
$$\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n - 1} \cdot 5^{2 n - 1}}{2 n - 1}$$
Sum((-1)^(-1 + n)*5^(-1 + 2*n)/(-1 + 2*n), (n, 0, oo))
Numerical answer
The series diverges
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