Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
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- Product of series
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How to use it?
- Sum of series:
-
sqrt((n+4)/((n^4)+4))
-
1/((3*n+1)*(3*n+4))
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lnn/n
-
sin(pi/2^(n-1))
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Identical expressions
- sin((two n- one)-x)/(2n- one)^2
- sinus of ((2n minus 1) minus x) divide by (2n minus 1) squared
- sinus of ((two n minus one) minus x) divide by (2n minus one) squared
- sin((2n-1)-x)/(2n-1)2
- sin2n-1-x/2n-12
- sin((2n-1)-x)/(2n-1)²
- sin((2n-1)-x)/(2n-1) to the power of 2
- sin2n-1-x/2n-1^2
- sin((2n-1)-x) divide by (2n-1)^2
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Similar expressions
- sin((2n-1)-x)/(2n+1)^2
- sin((2n+1)-x)/(2n-1)^2
- sin((2n-1)+x)/(2n-1)^2
Sum of series sin((2n-1)-x)/(2n-1)^2
The solution
You have entered
[src]
oo ____ \ ` \ sin(2*n - 1 - x) \ ---------------- / 2 / (2*n - 1) /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\sin{\left(- x + \left(2 n - 1\right) \right)}}{\left(2 n - 1\right)^{2}}$$
Sum(sin(2*n - 1 - x)/(2*n - 1)^2, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\sin{\left(- x + \left(2 n - 1\right) \right)}}{\left(2 n - 1\right)^{2}}$$
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
$$a_{n} = - \frac{\sin{\left(- 2 n + x + 1 \right)}}{\left(2 n - 1\right)^{2}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(2 n + 1\right)^{2} \left|{\frac{\sin{\left(- 2 n + x + 1 \right)}}{\left(2 n - 1\right)^{2} \sin{\left(2 n - x + 1 \right)}}}\right|\right)$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty}\left(\left(2 n + 1\right)^{2} \left|{\frac{\sin{\left(- 2 n + x + 1 \right)}}{\left(2 n - 1\right)^{2} \sin{\left(2 n - x + 1 \right)}}}\right|\right)$$
$$\frac{\sin{\left(- x + \left(2 n - 1\right) \right)}}{\left(2 n - 1\right)^{2}}$$
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
$$a_{n} = - \frac{\sin{\left(- 2 n + x + 1 \right)}}{\left(2 n - 1\right)^{2}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(2 n + 1\right)^{2} \left|{\frac{\sin{\left(- 2 n + x + 1 \right)}}{\left(2 n - 1\right)^{2} \sin{\left(2 n - x + 1 \right)}}}\right|\right)$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty}\left(\left(2 n + 1\right)^{2} \left|{\frac{\sin{\left(- 2 n + x + 1 \right)}}{\left(2 n - 1\right)^{2} \sin{\left(2 n - x + 1 \right)}}}\right|\right)$$
False
The answer
[src]
oo ____ \ ` \ -sin(1 + x - 2*n) \ ------------------ / 2 / (-1 + 2*n) /___, n = 1
$$\sum_{n=1}^{\infty} - \frac{\sin{\left(- 2 n + x + 1 \right)}}{\left(2 n - 1\right)^{2}}$$
Sum(-sin(1 + x - 2*n)/(-1 + 2*n)^2, (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