Mister Exam
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How to use it?
- Sum of series:
-
38
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arctan(n)/(n!)^2
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arctg(5/n)/n!
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ln((n^2+n+3)/(n^2+n+2))
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Identical expressions
- (- one)^(x+ one)/(2x- one)
- ( minus 1) to the power of (x plus 1) divide by (2x minus 1)
- ( minus one) to the power of (x plus one) divide by (2x minus one)
- (-1)(x+1)/(2x-1)
- -1x+1/2x-1
- -1^x+1/2x-1
- (-1)^(x+1) divide by (2x-1)
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Similar expressions
- (-1)^(x+1)/(2x+1)
- (-1)^(x-1)/(2x-1)
- (1)^(x+1)/(2x-1)
Sum of series (-1)^(x+1)/(2x-1)
The solution
You have entered
[src]
oo ____ \ ` \ x + 1 \ (-1) / --------- / 2*x - 1 /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\left(-1\right)^{x + 1}}{2 x - 1}$$
Sum((-1)^(x + 1)/(2*x - 1), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\left(-1\right)^{x + 1}}{2 x - 1}$$
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
$$a_{n} = \frac{\left(-1\right)^{x + 1}}{2 x - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} 1$$
Let's take the limit
we find
$$\frac{\left(-1\right)^{x + 1}}{2 x - 1}$$
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
$$a_{n} = \frac{\left(-1\right)^{x + 1}}{2 x - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} 1$$
Let's take the limit
we find
True
False
The answer
[src]
1 + x oo*(-1) ------------ -1 + 2*x
$$\frac{\infty \left(-1\right)^{x + 1}}{2 x - 1}$$
oo*(-1)^(1 + x)/(-1 + 2*x)
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