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:
-
(4*9^n-2^n)/18^n
-
1/((3*n+1)*(3*n+4))
- sqrt(n)*(x+3)^n/n^2+1
-
ln((n(n+2))/(n+1)^2)
-
Identical expressions
- sin*(2n+ one /(n^ three))
- sinus of multiply by (2n plus 1 divide by (n cubed ))
- sinus of multiply by (2n plus one divide by (n to the power of three))
- sin*(2n+1/(n3))
- sin*2n+1/n3
- sin*(2n+1/(n³))
- sin*(2n+1/(n to the power of 3))
- sin(2n+1/(n^3))
- sin(2n+1/(n3))
- sin2n+1/n3
- sin2n+1/n^3
- sin*(2n+1 divide by (n^3))
-
Similar expressions
- sin*(2n-1/(n^3))
Sum of series sin*(2n+1/(n^3))
The solution
You have entered
[src]
oo ____ \ ` \ / 1 \ \ sin|2*n + --| / | 3| / \ n / /___, n = 1
$$\sum_{n=1}^{\infty} \sin{\left(2 n + \frac{1}{n^{3}} \right)}$$
Sum(sin(2*n + 1/(n^3)), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\sin{\left(2 n + \frac{1}{n^{3}} \right)}$$
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
$$a_{n} = \sin{\left(2 n + \frac{1}{n^{3}} \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\sin{\left(2 n + \frac{1}{n^{3}} \right)}}{\sin{\left(2 n + 2 + \frac{1}{\left(n + 1\right)^{3}} \right)}}}\right|$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty} \left|{\frac{\sin{\left(2 n + \frac{1}{n^{3}} \right)}}{\sin{\left(2 n + 2 + \frac{1}{\left(n + 1\right)^{3}} \right)}}}\right|$$
$$\sin{\left(2 n + \frac{1}{n^{3}} \right)}$$
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
$$a_{n} = \sin{\left(2 n + \frac{1}{n^{3}} \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\sin{\left(2 n + \frac{1}{n^{3}} \right)}}{\sin{\left(2 n + 2 + \frac{1}{\left(n + 1\right)^{3}} \right)}}}\right|$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty} \left|{\frac{\sin{\left(2 n + \frac{1}{n^{3}} \right)}}{\sin{\left(2 n + 2 + \frac{1}{\left(n + 1\right)^{3}} \right)}}}\right|$$
False
The rate of convergence of the power series
The answer
[src]
oo ____ \ ` \ /1 \ \ sin|-- + 2*n| / | 3 | / \n / /___, n = 1
$$\sum_{n=1}^{\infty} \sin{\left(2 n + \frac{1}{n^{3}} \right)}$$
Sum(sin(n^(-3) + 2*n), (n, 1, oo))
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