Mister Exam
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How to use it?
- Sum of series:
-
45^(n+4)/(3^(2*n+7)*5^(n+3))
-
1/4+1/28+1/70
- ((-1)^n)*x^2n/(2n+1)!
- ln(k+1)/n
-
Identical expressions
- forty-five ^(n+ four)/(three ^(two *n+ seven)* five ^(n+ three))
- 45 to the power of (n plus 4) divide by (3 to the power of (2 multiply by n plus 7) multiply by 5 to the power of (n plus 3))
- forty minus five to the power of (n plus four) divide by (three to the power of (two multiply by n plus seven) multiply by five to the power of (n plus three))
- 45(n+4)/(3(2*n+7)*5(n+3))
- 45n+4/32*n+7*5n+3
- 45^(n+4)/(3^(2n+7)5^(n+3))
- 45(n+4)/(3(2n+7)5(n+3))
- 45n+4/32n+75n+3
- 45^n+4/3^2n+75^n+3
- 45^(n+4) divide by (3^(2*n+7)*5^(n+3))
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Similar expressions
- 45^(n+4)/(3^(2*n-7)*5^(n+3))
- 45^(n+4)/(3^(2*n+7)*5^(n-3))
- 45^(n-4)/(3^(2*n+7)*5^(n+3))
Sum of series 45^(n+4)/(3^(2*n+7)*5^(n+3))
The solution
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oo ____ \ ` \ n + 4 \ 45 ) --------------- / 2*n + 7 n + 3 / 3 *5 /___, n = 1
$$\sum_{n=1}^{\infty} \frac{45^{n + 4}}{3^{2 n + 7} \cdot 5^{n + 3}}$$
Sum(45^(n + 4)/((3^(2*n + 7)*5^(n + 3))), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{45^{n + 4}}{3^{2 n + 7} \cdot 5^{n + 3}}$$
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
$$a_{n} = 3^{- 2 n - 7} \cdot 45^{n + 4} \cdot 5^{- n - 3}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(225^{n + 4} \cdot 3^{- 2 n - 7} \cdot 3^{2 n + 9} \cdot 45^{- n - 5} \cdot 5^{- n - 3}\right)$$
Let's take the limit
we find
$$\frac{45^{n + 4}}{3^{2 n + 7} \cdot 5^{n + 3}}$$
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
$$a_{n} = 3^{- 2 n - 7} \cdot 45^{n + 4} \cdot 5^{- n - 3}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(225^{n + 4} \cdot 3^{- 2 n - 7} \cdot 3^{2 n + 9} \cdot 45^{- n - 5} \cdot 5^{- n - 3}\right)$$
Let's take the limit
we find
True
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