Mister Exam
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- Taylor Series Step by Step
- Fourier series step by step
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- Product of series
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How to use it?
- Sum of series:
-
(n^3+n+5)/(n+2)
-
(0.15/0.05)^1.3
-
5/(25*n^2-5*n-6)
-
tan(sqrt(n)/(n^2+1))
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Identical expressions
- -(sixteen / two ^n)*n
- minus (16 divide by 2 to the power of n) multiply by n
- minus (sixteen divide by two to the power of n) multiply by n
- -(16/2n)*n
- -16/2n*n
- -(16/2^n)n
- -(16/2n)n
- -16/2nn
- -16/2^nn
- -(16 divide by 2^n)*n
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Similar expressions
- (16/2^n)*n
Sum of series -(16/2^n)*n
The solution
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[src]
oo ___ \ ` \ n / -8 *n /__, n = 1
$$\sum_{n=1}^{\infty} - 8^{n} n$$
Sum((-8^n)*n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$- 8^{n} n$$
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
$$a_{n} = - n$$
and
$$x_{0} = -8$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(-8 + \lim_{n \to \infty}\left(\frac{n}{n + 1}\right)\right)$$
Let's take the limit
we find
$$- 8^{n} n$$
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
$$a_{n} = - n$$
and
$$x_{0} = -8$$
,
$$d = 1$$
,
$$c = 0$$
then
$$R = \tilde{\infty} \left(-8 + \lim_{n \to \infty}\left(\frac{n}{n + 1}\right)\right)$$
Let's take the limit
we find
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