Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
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- Product of series
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How to use it?
- Sum of series:
- tg^(2)*(6/(2n)^(1/3))*(x-24)^n
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1/(2*n-1)*(2*n+1)
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((n+1)^n*3^(2n-1))/(2n+1)!
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ln^2n*2
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Identical expressions
- ((n+ one)^n* three ^(2n- one))/(2n+ one)!
- ((n plus 1) to the power of n multiply by 3 to the power of (2n minus 1)) divide by (2n plus 1)!
- ((n plus one) to the power of n multiply by three to the power of (2n minus one)) divide by (2n plus one)!
- ((n+1)n*3(2n-1))/(2n+1)!
- n+1n*32n-1/2n+1!
- ((n+1)^n3^(2n-1))/(2n+1)!
- ((n+1)n3(2n-1))/(2n+1)!
- n+1n32n-1/2n+1!
- n+1^n3^2n-1/2n+1!
- ((n+1)^n*3^(2n-1)) divide by (2n+1)!
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Similar expressions
- ((n-1)^n*3^(2n-1))/(2n+1)!
- ((n+1)^n*3^(2n-1))/(2n-1)!
- ((n+1)^n*3^(2n+1))/(2n+1)!
Sum of series ((n+1)^n*3^(2n-1))/(2n+1)!
The solution
You have entered
[src]
oo ____ \ ` \ n 2*n - 1 \ (n + 1) *3 / ----------------- / (2*n + 1)! /___, n = 1
$$\sum_{n=1}^{\infty} \frac{3^{2 n - 1} \left(n + 1\right)^{n}}{\left(2 n + 1\right)!}$$
Sum(((n + 1)^n*3^(2*n - 1))/factorial(2*n + 1), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{3^{2 n - 1} \left(n + 1\right)^{n}}{\left(2 n + 1\right)!}$$
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
$$a_{n} = \frac{3^{2 n - 1} \left(n + 1\right)^{n}}{\left(2 n + 1\right)!}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(3^{- 2 n - 1} \cdot 3^{2 n - 1} \left(n + 1\right)^{n} \left(n + 2\right)^{- n - 1} \left|{\frac{\left(2 n + 3\right)!}{\left(2 n + 1\right)!}}\right|\right)$$
Let's take the limit
we find
$$\frac{3^{2 n - 1} \left(n + 1\right)^{n}}{\left(2 n + 1\right)!}$$
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
$$a_{n} = \frac{3^{2 n - 1} \left(n + 1\right)^{n}}{\left(2 n + 1\right)!}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(3^{- 2 n - 1} \cdot 3^{2 n - 1} \left(n + 1\right)^{n} \left(n + 2\right)^{- n - 1} \left|{\frac{\left(2 n + 3\right)!}{\left(2 n + 1\right)!}}\right|\right)$$
Let's take the limit
we find
False
False
The rate of convergence of the power series
The answer
[src]
oo ____ \ ` \ -1 + 2*n n \ 3 *(1 + n) / ------------------ / (1 + 2*n)! /___, n = 1
$$\sum_{n=1}^{\infty} \frac{3^{2 n - 1} \left(n + 1\right)^{n}}{\left(2 n + 1\right)!}$$
Sum(3^(-1 + 2*n)*(1 + n)^n/factorial(1 + 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