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:
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24/(9n^2-12n-5)
-
n^(2/3)*arctg(1/n^2)
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2^(2*n-1)/9^n
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1/4^1
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Identical expressions
- twenty-four /(9n^ two -12n- five)
- 24 divide by (9n squared minus 12n minus 5)
- twenty minus four divide by (9n to the power of two minus 12n minus five)
- 24/(9n2-12n-5)
- 24/9n2-12n-5
- 24/(9n²-12n-5)
- 24/(9n to the power of 2-12n-5)
- 24/9n^2-12n-5
- 24 divide by (9n^2-12n-5)
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Similar expressions
- 24/(9n^2+12n-5)
- 24/(9n^2-12n+5)
Sum of series 24/(9n^2-12n-5)
The solution
You have entered
[src]
oo ____ \ ` \ 24 \ --------------- / 2 / 9*n - 12*n - 5 /___, n = 1
$$\sum_{n=1}^{\infty} \frac{24}{\left(9 n^{2} - 12 n\right) - 5}$$
Sum(24/(9*n^2 - 12*n - 5), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{24}{\left(9 n^{2} - 12 n\right) - 5}$$
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
$$a_{n} = \frac{24}{9 n^{2} - 12 n - 5}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(24 \left|{\frac{\frac{n}{2} - \frac{3 \left(n + 1\right)^{2}}{8} + \frac{17}{24}}{- 9 n^{2} + 12 n + 5}}\right|\right)$$
Let's take the limit
we find
$$\frac{24}{\left(9 n^{2} - 12 n\right) - 5}$$
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
$$a_{n} = \frac{24}{9 n^{2} - 12 n - 5}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(24 \left|{\frac{\frac{n}{2} - \frac{3 \left(n + 1\right)^{2}}{8} + \frac{17}{24}}{- 9 n^{2} + 12 n + 5}}\right|\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
The answer
[src]
3*Gamma(7/3) ------------ 2*Gamma(4/3)
$$\frac{3 \Gamma\left(\frac{7}{3}\right)}{2 \Gamma\left(\frac{4}{3}\right)}$$
3*gamma(7/3)/(2*gamma(4/3))
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