Mister Exam
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- Product of series
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How to use it?
- Sum of series:
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1/(n(n+2))
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1/n^6
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(3^n-4^n)/12^n
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2i
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Identical expressions
- (seven ^(n+ one))(eight ^-n)
- (7 to the power of (n plus 1))(8 to the power of minus n)
- (seven to the power of (n plus one))(eight to the power of minus n)
- (7(n+1))(8-n)
- 7n+18-n
- 7^n+18^-n
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Similar expressions
- (7^(n-1))(8^-n)
- (7^(n+1))(8^+n)
Sum of series (7^(n+1))(8^-n)
The solution
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oo ___ \ ` \ n + 1 -n / 7 *8 /__, n = 1
$$\sum_{n=1}^{\infty} 7^{n + 1} \cdot 8^{- n}$$
Sum(7^(n + 1)*8^(-n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$7^{n + 1} \cdot 8^{- n}$$
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
$$a_{n} = 7^{n + 1}$$
and
$$x_{0} = -8$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-8 + \lim_{n \to \infty}\left(7^{- n - 2} \cdot 7^{n + 1}\right)\right)$$
Let's take the limit
we find
$$R = 0$$
$$7^{n + 1} \cdot 8^{- n}$$
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
$$a_{n} = 7^{n + 1}$$
and
$$x_{0} = -8$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-8 + \lim_{n \to \infty}\left(7^{- n - 2} \cdot 7^{n + 1}\right)\right)$$
Let's take the limit
we find
False
False
$$R = 0$$
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