Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
- Differential equations Step by Step
- Product of series
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How to use it?
- Sum of series:
-
1/(2n+1)
-
1+1/factorial(n)
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(3n+4)/(n/((n+1)(n+2)))
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((-1)^n)(pi^(2n))/((6^(2n))(2n)!)
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Identical expressions
- (3n+ four)/(n/((n+ one)(n+ two)))
- (3n plus 4) divide by (n divide by ((n plus 1)(n plus 2)))
- (3n plus four) divide by (n divide by ((n plus one)(n plus two)))
- 3n+4/n/n+1n+2
- (3n+4) divide by (n divide by ((n+1)(n+2)))
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Similar expressions
- (3n+4)/(n/(n+1)(n+2))
- (3n+4)/(n/((n+1)(n-2)))
- (3n-4)/(n/((n+1)(n+2)))
- (3n+4)/(n/((n-1)(n+2)))
Sum of series (3n+4)/(n/((n+1)(n+2)))
The solution
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oo ____ \ ` \ 3*n + 4 \ ----------------- ) / n \ / |---------------| / \(n + 1)*(n + 2)/ /___, n = 1
$$\sum_{n=1}^{\infty} \frac{3 n + 4}{n \frac{1}{\left(n + 1\right) \left(n + 2\right)}}$$
Sum((3*n + 4)/((n/(((n + 1)*(n + 2))))), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{3 n + 4}{n \frac{1}{\left(n + 1\right) \left(n + 2\right)}}$$
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
$$a_{n} = \frac{\left(n + 1\right) \left(n + 2\right) \left(3 n + 4\right)}{n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2} \left(3 n + 4\right)}{n \left(n + 3\right) \left(3 n + 7\right)}\right)$$
Let's take the limit
we find
$$\frac{3 n + 4}{n \frac{1}{\left(n + 1\right) \left(n + 2\right)}}$$
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
$$a_{n} = \frac{\left(n + 1\right) \left(n + 2\right) \left(3 n + 4\right)}{n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right)^{2} \left(3 n + 4\right)}{n \left(n + 3\right) \left(3 n + 7\right)}\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
Numerical answer
The series diverges
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