Mister Exam
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- Fourier series step by step
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- Product of series
-
How to use it?
- Sum of series:
-
1/(n(n+2))
-
(2n-1)/2^n
-
sin(3n)/(7n)^(1/5)
-
arctg1/(2n^2)
-
Identical expressions
- arctg1/(two n^2)
- arctg1 divide by (2n squared )
- arctg1 divide by (two n squared )
- arctg1/(2n2)
- arctg1/2n2
- arctg1/(2n²)
- arctg1/(2n to the power of 2)
- arctg1/2n^2
- arctg1 divide by (2n^2)
Sum of series arctg1/(2n^2)
The solution
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oo ____ \ ` \ atan(1) \ ------- / 2 / 2*n /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\operatorname{atan}{\left(1 \right)}}{2 n^{2}}$$
Sum(atan(1)/((2*n^2)), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\operatorname{atan}{\left(1 \right)}}{2 n^{2}}$$
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
$$a_{n} = \frac{\pi}{8 n^{2}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)$$
Let's take the limit
we find
$$\frac{\operatorname{atan}{\left(1 \right)}}{2 n^{2}}$$
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
$$a_{n} = \frac{\pi}{8 n^{2}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2}}{n^{2}}\right)$$
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