Mister Exam
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How to use it?
- Sum of series:
-
(3/4)^n
-
1/((2n-1)*(2n+1))
-
1/(n*(n+3))
-
log(1-1/n^2)
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Identical expressions
- eight ^(n- one)/(n- one)!
- 8 to the power of (n minus 1) divide by (n minus 1)!
- eight to the power of (n minus one) divide by (n minus one)!
- 8(n-1)/(n-1)!
- 8n-1/n-1!
- 8^n-1/n-1!
- 8^(n-1) divide by (n-1)!
-
Similar expressions
- 8^(n+1)/(n-1)!
- 8^(n-1)/(n+1)!
Sum of series 8^(n-1)/(n-1)!
The solution
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oo ____ \ ` \ n - 1 \ 8 / -------- / (n - 1)! /___, n = 1
$$\sum_{n=1}^{\infty} \frac{8^{n - 1}}{\left(n - 1\right)!}$$
Sum(8^(n - 1)/factorial(n - 1), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{8^{n - 1}}{\left(n - 1\right)!}$$
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
$$a_{n} = \frac{8^{n - 1}}{\left(n - 1\right)!}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(8^{- n} 8^{n - 1} \left|{\frac{n!}{\left(n - 1\right)!}}\right|\right)$$
Let's take the limit
we find
$$\frac{8^{n - 1}}{\left(n - 1\right)!}$$
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
$$a_{n} = \frac{8^{n - 1}}{\left(n - 1\right)!}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(8^{- n} 8^{n - 1} \left|{\frac{n!}{\left(n - 1\right)!}}\right|\right)$$
Let's take the limit
we find
False
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