Mister Exam
Other calculators
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- Product of series
-
How to use it?
- Sum of series:
- ((-1)^n)*((x^n)/(2n*n!))
-
sin²n/√n³+5
-
factorial(n+1)/8^(n+1)
-
x+5
-
Identical expressions
- lim(two n^2+ three)^ three / three n^ four -3
- lim(2n squared plus 3) cubed divide by 3n to the power of 4 minus 3
- lim(two n squared plus three) to the power of three divide by three n to the power of four minus 3
- lim(2n2+3)3/3n4-3
- lim2n2+33/3n4-3
- lim(2n²+3)³/3n⁴-3
- lim(2n to the power of 2+3) to the power of 3/3n to the power of 4-3
- lim2n^2+3^3/3n^4-3
- lim(2n^2+3)^3 divide by 3n^4-3
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Similar expressions
- lim(2n^2-3)^3/3n^4-3
- lim(2n^2+3)^3/3n^4+3
Sum of series lim(2n^2+3)^3/3n^4-3
The solution
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[src]
oo ____ \ ` \ / 3 \ \ |/ 2 \ | ) |\2*n + 3/ 4 | / |-----------*n - 3| / \ 3 / /___, n = 1
$$\sum_{n=1}^{\infty} \left(n^{4} \frac{\left(2 n^{2} + 3\right)^{3}}{3} - 3\right)$$
Sum(((2*n^2 + 3)^3/3)*n^4 - 3, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$n^{4} \frac{\left(2 n^{2} + 3\right)^{3}}{3} - 3$$
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
$$a_{n} = \frac{n^{4} \left(2 n^{2} + 3\right)^{3}}{3} - 3$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left|{\frac{n^{4} \left(2 n^{2} + 3\right)^{3}}{3} - 3}\right|}{\frac{\left(n + 1\right)^{4} \left(2 \left(n + 1\right)^{2} + 3\right)^{3}}{3} - 3}\right)$$
Let's take the limit
we find
$$n^{4} \frac{\left(2 n^{2} + 3\right)^{3}}{3} - 3$$
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
$$a_{n} = \frac{n^{4} \left(2 n^{2} + 3\right)^{3}}{3} - 3$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left|{\frac{n^{4} \left(2 n^{2} + 3\right)^{3}}{3} - 3}\right|}{\frac{\left(n + 1\right)^{4} \left(2 \left(n + 1\right)^{2} + 3\right)^{3}}{3} - 3}\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
The answer
[src]
oo ____ \ ` \ / 3\ \ | 4 / 2\ | ) | n *\3 + 2*n / | / |-3 + --------------| / \ 3 / /___, n = 1
$$\sum_{n=1}^{\infty} \left(\frac{n^{4} \left(2 n^{2} + 3\right)^{3}}{3} - 3\right)$$
Sum(-3 + n^4*(3 + 2*n^2)^3/3, (n, 1, oo))
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