Mister Exam
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- Product of series
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How to use it?
- Sum of series:
-
1/((2n-1)(2n+5))
-
1/(n*(n+2))
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1/5^n
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(5^n+4^n)/6^n
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Identical expressions
- one /((2n- one)(2n+ five))
- 1 divide by ((2n minus 1)(2n plus 5))
- one divide by ((2n minus one)(2n plus five))
- 1/2n-12n+5
- 1 divide by ((2n-1)(2n+5))
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Similar expressions
- 1/(2n-1)(2n+5)
- 1/(2*n-1)*(2*n+5)
- 1/((2n-1)*(2n+5))
- 1/(2*n-1)(2*n+5)
- 1/7+1/27+1/(2n-1)*(2n+5)
- 1/((2n+1)(2n+5))
- 1/((2n-1)(2n-5))
Sum of series 1/((2n-1)(2n+5))
The solution
You have entered
[src]
oo ___ \ ` \ 1 ) ------------------- / (2*n - 1)*(2*n + 5) /__, n = 1
$$\sum_{n=1}^{\infty} \frac{1}{\left(2 n - 1\right) \left(2 n + 5\right)}$$
Sum(1/((2*n - 1)*(2*n + 5)), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{\left(2 n - 1\right) \left(2 n + 5\right)}$$
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
$$a_{n} = \frac{1}{\left(2 n - 1\right) \left(2 n + 5\right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(2 n + 1\right) \left(2 n + 7\right) \left|{\frac{1}{2 n - 1}}\right|}{2 n + 5}\right)$$
Let's take the limit
we find
$$\frac{1}{\left(2 n - 1\right) \left(2 n + 5\right)}$$
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
$$a_{n} = \frac{1}{\left(2 n - 1\right) \left(2 n + 5\right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(2 n + 1\right) \left(2 n + 7\right) \left|{\frac{1}{2 n - 1}}\right|}{2 n + 5}\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
The answer
[src]
0 0 -1 + 5*e -14 + 8*e --------------- + --------------- / 0\ / 0\ 4*\-15 + 15*e / 6*\-15 + 15*e /
$$\frac{-14 + 8 e^{0}}{6 \left(-15 + 15 e^{0}\right)} + \frac{-1 + 5 e^{0}}{4 \left(-15 + 15 e^{0}\right)}$$
(-1 + 5*exp_polar(0))/(4*(-15 + 15*exp_polar(0))) + (-14 + 8*exp_polar(0))/(6*(-15 + 15*exp_polar(0)))
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