Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
- Differential equations Step by Step
- Product of series
-
How to use it?
- Sum of series:
-
(8/((2*n+1)*(2*n+1)*3.1416*3.1416))*exp(-(2*n+1)*0.197*3.1416*3.1416*0.25)
- factorial(k)/((factorial(n)*factorial(n+k)))
-
0.5^n
-
1/(3n+2)(3n-1)
-
Identical expressions
- sin(2nx)*(- one)^(n+ one)/n
- sinus of (2nx) multiply by ( minus 1) to the power of (n plus 1) divide by n
- sinus of (2nx) multiply by ( minus one) to the power of (n plus one) divide by n
- sin(2nx)*(-1)(n+1)/n
- sin2nx*-1n+1/n
- sin(2nx)(-1)^(n+1)/n
- sin(2nx)(-1)(n+1)/n
- sin2nx-1n+1/n
- sin2nx-1^n+1/n
- sin(2nx)*(-1)^(n+1) divide by n
-
Similar expressions
- sin(2nx)*(-1)^(n-1)/n
- sin(2nx)*(1)^(n+1)/n
Sum of series sin(2nx)*(-1)^(n+1)/n
The solution
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[src]
oo ____ \ ` \ n + 1 \ sin(2*n*x)*(-1) / -------------------- / n /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1} \sin{\left(2 n x \right)}}{n}$$
Sum((sin((2*n)*x)*(-1)^(n + 1))/n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\left(-1\right)^{n + 1} \sin{\left(2 n x \right)}}{n}$$
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
$$a_{n} = \frac{\left(-1\right)^{n + 1} \sin{\left(2 n x \right)}}{n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left|{\frac{\sin{\left(2 n x \right)}}{\sin{\left(2 x \left(n + 1\right) \right)}}}\right|}{n}\right)$$
Let's take the limit
we find
$$\frac{\left(-1\right)^{n + 1} \sin{\left(2 n x \right)}}{n}$$
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
$$a_{n} = \frac{\left(-1\right)^{n + 1} \sin{\left(2 n x \right)}}{n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left|{\frac{\sin{\left(2 n x \right)}}{\sin{\left(2 x \left(n + 1\right) \right)}}}\right|}{n}\right)$$
Let's take the limit
we find
True
False
The answer
[src]
oo ____ \ ` \ 1 + n \ (-1) *sin(2*n*x) / -------------------- / n /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\left(-1\right)^{n + 1} \sin{\left(2 n x \right)}}{n}$$
Sum((-1)^(1 + n)*sin(2*n*x)/n, (n, 1, oo))
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