Mister Exam
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How to use it?
- Sum of series:
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25
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12(0.73)^(n-1)
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1/(n*lnn)
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(-1)^(n+1)/log(n+2^0.5)
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Identical expressions
- twelve (zero . seventy-three)^(n- one)
- 12(0.73) to the power of (n minus 1)
- twelve (zero . seventy minus three) to the power of (n minus one)
- 12(0.73)(n-1)
- 120.73n-1
- 120.73^n-1
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Similar expressions
- 12(0.73)^(n+1)
Sum of series 12(0.73)^(n-1)
The solution
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oo ____ \ ` \ n - 1 \ / 73\ / 12*|---| / \100/ /___, n = 1
$$\sum_{n=1}^{\infty} 12 \left(\frac{73}{100}\right)^{n - 1}$$
Sum(12*(73/100)^(n - 1), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$12 \left(\frac{73}{100}\right)^{n - 1}$$
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
$$a_{n} = 12 \left(\frac{73}{100}\right)^{n - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(\frac{73}{100}\right)^{- n} \left(\frac{73}{100}\right)^{n - 1}\right)$$
Let's take the limit
we find
$$12 \left(\frac{73}{100}\right)^{n - 1}$$
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
$$a_{n} = 12 \left(\frac{73}{100}\right)^{n - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(\frac{73}{100}\right)^{- n} \left(\frac{73}{100}\right)^{n - 1}\right)$$
Let's take the limit
we find
False
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