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:
-
1/2^(2n-2)
-
(6+(-4)^n)/(7^(n+2))
-
x²-16
-
ln(1-(1/(k)^2))
-
Identical expressions
- one / two ^(two n-2)
- 1 divide by 2 to the power of (2n minus 2)
- one divide by two to the power of (two n minus 2)
- 1/2(2n-2)
- 1/22n-2
- 1/2^2n-2
- 1 divide by 2^(2n-2)
-
Similar expressions
- 1/2^(2n+2)
Sum of series 1/2^(2n-2)
The solution
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[src]
oo ___ \ ` \ 2 - 2*n / 2 /__, n = 1
$$\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{2 n - 2}$$
Sum((1/2)^(2*n - 2), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\left(\frac{1}{2}\right)^{2 n - 2}$$
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
$$a_{n} = \left(\frac{1}{2}\right)^{2 n - 2}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(2^{2 n} 2^{2 - 2 n}\right)$$
Let's take the limit
we find
$$\left(\frac{1}{2}\right)^{2 n - 2}$$
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
$$a_{n} = \left(\frac{1}{2}\right)^{2 n - 2}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(2^{2 n} 2^{2 - 2 n}\right)$$
Let's take the limit
we find
False
False
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