Mister Exam
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- Product of series
-
How to use it?
- Sum of series:
-
((-1)^(n+1))/(2^n)
-
5/n
- (-1)^n*x^n/(2*n+1)
- (x^n)/(n*(n+1))
-
Identical expressions
- ((- one)^(n+ one))/(two ^n)
- (( minus 1) to the power of (n plus 1)) divide by (2 to the power of n)
- (( minus one) to the power of (n plus one)) divide by (two to the power of n)
- ((-1)(n+1))/(2n)
- -1n+1/2n
- -1^n+1/2^n
- ((-1)^(n+1)) divide by (2^n)
-
Similar expressions
- ((-1)^(n-1))/(2^n)
- ((1)^(n+1))/(2^n)
Sum of series ((-1)^(n+1))/(2^n)
The solution
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oo ____ \ ` \ n + 1 \ (-1) ) --------- / n / 2 /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1}}{2^{n}}$$
Sum((-1)^(n + 1)/2^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\left(-1\right)^{n + 1}}{2^{n}}$$
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
$$a_{n} = \left(-1\right)^{n + 1}$$
and
$$x_{0} = -2$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} 1\right)$$
Let's take the limit
we find
$$R = 0$$
$$\frac{\left(-1\right)^{n + 1}}{2^{n}}$$
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
$$a_{n} = \left(-1\right)^{n + 1}$$
and
$$x_{0} = -2$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-2 + \lim_{n \to \infty} 1\right)$$
Let's take the limit
we find
False
$$R = 0$$
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