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:
-
1/(1*3)+1/((2n-1)*(2n+1))
- ((-1)^n*(3*cos(n*x)-pi*n*sin(n*x))-3*cos(n*x+pi*n/3))/n^2
- 7^x/(5^x-3)
-
(5-((2^n)-1))/(5-(2^n))
-
Identical expressions
- sqrt(n/n^ three + two *n+ nine)
- square root of (n divide by n cubed plus 2 multiply by n plus 9)
- square root of (n divide by n to the power of three plus two multiply by n plus nine)
- √(n/n^3+2*n+9)
- sqrt(n/n3+2*n+9)
- sqrtn/n3+2*n+9
- sqrt(n/n³+2*n+9)
- sqrt(n/n to the power of 3+2*n+9)
- sqrt(n/n^3+2n+9)
- sqrt(n/n3+2n+9)
- sqrtn/n3+2n+9
- sqrtn/n^3+2n+9
- sqrt(n divide by n^3+2*n+9)
-
Similar expressions
- sqrt(n/n^3+2*n-9)
- sqrt(n/n^3-2*n+9)
Sum of series sqrt(n/n^3+2*n+9)
The solution
You have entered
[src]
oo ____ \ ` \ ______________ \ / n ) / -- + 2*n + 9 / / 3 / \/ n /___, n = 1
$$\sum_{n=1}^{\infty} \sqrt{\left(2 n + \frac{n}{n^{3}}\right) + 9}$$
Sum(sqrt(n/n^3 + 2*n + 9), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\sqrt{\left(2 n + \frac{n}{n^{3}}\right) + 9}$$
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
$$a_{n} = \sqrt{2 n + 9 + \frac{1}{n^{2}}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt{2 n + 9 + \frac{1}{n^{2}}}}{\sqrt{2 n + 11 + \frac{1}{\left(n + 1\right)^{2}}}}\right)$$
Let's take the limit
we find
$$\sqrt{\left(2 n + \frac{n}{n^{3}}\right) + 9}$$
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
$$a_{n} = \sqrt{2 n + 9 + \frac{1}{n^{2}}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt{2 n + 9 + \frac{1}{n^{2}}}}{\sqrt{2 n + 11 + \frac{1}{\left(n + 1\right)^{2}}}}\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
The answer
[src]
oo ____ \ ` \ ______________ \ / 1 ) / 9 + -- + 2*n / / 2 / \/ n /___, n = 1
$$\sum_{n=1}^{\infty} \sqrt{2 n + 9 + \frac{1}{n^{2}}}$$
Sum(sqrt(9 + n^(-2) + 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