Mister Exam
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- Taylor Series Step by Step
- Fourier series step by step
- Differential equations Step by Step
- Product of series
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How to use it?
- Sum of series:
-
(7^n-2^n)/14^n
-
3/n
-
log(1+3/n)
- (-1)^(x+1)/(2x-1)
-
Identical expressions
- (four)/4n- three *4n+ one
- (4) divide by 4n minus 3 multiply by 4n plus 1
- (four) divide by 4n minus three multiply by 4n plus one
- (4)/4n-34n+1
- 4/4n-34n+1
- (4) divide by 4n-3*4n+1
-
Similar expressions
- (4)/4n+3*4n+1
- (4)/4n-3*4n-1
- 4/((4n-3)(4n+1))
Sum of series (4)/4n-3*4n+1
The solution
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oo __ \ ` ) (1.0*n - 12*n + 1) /_, n = 1
$$\sum_{n=1}^{\infty} \left(\left(- 12 n + 1.0 n\right) + 1\right)$$
Sum(1.0*n - 12*n + 1, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\left(- 12 n + 1 n\right) + 1$$
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
$$a_{n} = 1 - 11 n$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left|{11 n - 1}\right|}{11 n + 10}\right)$$
Let's take the limit
we find
$$\left(- 12 n + 1 n\right) + 1$$
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
$$a_{n} = 1 - 11 n$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left|{11 n - 1}\right|}{11 n + 10}\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
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