Mister Exam
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- Product of series
-
How to use it?
- Sum of series:
-
5/pi^n
-
1/(4n^(2)-1)
-
(n+1)/factorial(n)
-
13n(13n+13)-(13n-13)(13+13n)
-
Identical expressions
- one /(4n^(two)- one)
- 1 divide by (4n to the power of (2) minus 1)
- one divide by (4n to the power of (two) minus one)
- 1/(4n(2)-1)
- 1/4n2-1
- 1/4n^2-1
- 1 divide by (4n^(2)-1)
-
Similar expressions
- 1/(4n^(2)+1)
Sum of series 1/(4n^(2)-1)
The solution
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oo ____ \ ` \ 1 \ -------- / 2 / 4*n - 1 /___, n = 0
$$\sum_{n=0}^{\infty} \frac{1}{4 n^{2} - 1}$$
Sum(1/(4*n^2 - 1), (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{4 n^{2} - 1}$$
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
$$a_{n} = \frac{1}{4 n^{2} - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(4 \left(n + 1\right)^{2} - 1\right) \left|{\frac{1}{4 n^{2} - 1}}\right|\right)$$
Let's take the limit
we find
$$\frac{1}{4 n^{2} - 1}$$
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
$$a_{n} = \frac{1}{4 n^{2} - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(4 \left(n + 1\right)^{2} - 1\right) \left|{\frac{1}{4 n^{2} - 1}}\right|\right)$$
Let's take the limit
we find
True
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