Mister Exam
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How to use it?
- Sum of series:
- n^2*x^n
-
9/(9n^2+21n-8)
-
(n^3+n+5)/(n+6)
- sen(ix)
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Identical expressions
- arctg(x^ three)/(n(n+ two)(n+ three))
- arctg(x cubed ) divide by (n(n plus 2)(n plus 3))
- arctg(x to the power of three) divide by (n(n plus two)(n plus three))
- arctg(x3)/(n(n+2)(n+3))
- arctgx3/nn+2n+3
- arctg(x³)/(n(n+2)(n+3))
- arctg(x to the power of 3)/(n(n+2)(n+3))
- arctgx^3/nn+2n+3
- arctg(x^3) divide by (n(n+2)(n+3))
-
Similar expressions
- arctg(x^3)/(n(n-2)(n+3))
- arctg(x^3)/(n(n+2)(n-3))
Sum of series arctg(x^3)/(n(n+2)(n+3))
The solution
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oo ____ \ ` \ / 3\ \ atan\x / / ----------------- / n*(n + 2)*(n + 3) /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\operatorname{atan}{\left(x^{3} \right)}}{n \left(n + 2\right) \left(n + 3\right)}$$
Sum(atan(x^3)/(((n*(n + 2))*(n + 3))), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\operatorname{atan}{\left(x^{3} \right)}}{n \left(n + 2\right) \left(n + 3\right)}$$
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
$$a_{n} = \frac{\operatorname{atan}{\left(x^{3} \right)}}{n \left(n + 2\right) \left(n + 3\right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 2\right)}\right)$$
Let's take the limit
we find
$$\frac{\operatorname{atan}{\left(x^{3} \right)}}{n \left(n + 2\right) \left(n + 3\right)}$$
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
$$a_{n} = \frac{\operatorname{atan}{\left(x^{3} \right)}}{n \left(n + 2\right) \left(n + 3\right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left(n + 4\right)}{n \left(n + 2\right)}\right)$$
Let's take the limit
we find
True
False
The answer
[src]
/ 3\
5*atan\x /
----------
36
$$\frac{5 \operatorname{atan}{\left(x^{3} \right)}}{36}$$
5*atan(x^3)/36
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