Mister Exam
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- Product of series
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How to use it?
- Sum of series:
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sqrt(n)*(n/(4*n-3))^(2*n)
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100
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1/(n*ln(n)*ln(ln(n)))
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2
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Identical expressions
- sqrt(n)*(n/(four *n- three))^(two *n)
- square root of (n) multiply by (n divide by (4 multiply by n minus 3)) to the power of (2 multiply by n)
- square root of (n) multiply by (n divide by (four multiply by n minus three)) to the power of (two multiply by n)
- √(n)*(n/(4*n-3))^(2*n)
- sqrt(n)*(n/(4*n-3))(2*n)
- sqrtn*n/4*n-32*n
- sqrt(n)(n/(4n-3))^(2n)
- sqrt(n)(n/(4n-3))(2n)
- sqrtnn/4n-32n
- sqrtnn/4n-3^2n
- sqrt(n)*(n divide by (4*n-3))^(2*n)
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Similar expressions
- sqrt(n)*(n/(4*n+3))^(2*n)
Sum of series sqrt(n)*(n/(4*n-3))^(2*n)
The solution
You have entered
[src]
oo ____ \ ` \ 2*n \ ___ / n \ / \/ n *|-------| / \4*n - 3/ /___, n = 1
$$\sum_{n=1}^{\infty} \sqrt{n} \left(\frac{n}{4 n - 3}\right)^{2 n}$$
Sum(sqrt(n)*(n/(4*n - 3))^(2*n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\sqrt{n} \left(\frac{n}{4 n - 3}\right)^{2 n}$$
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
$$a_{n} = \sqrt{n} \left(\frac{n}{4 n - 3}\right)^{2 n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt{n} \left(\frac{n + 1}{4 n + 1}\right)^{- 2 n - 2} \left|{\left(\frac{n}{4 n - 3}\right)^{2 n}}\right|}{\sqrt{n + 1}}\right)$$
Let's take the limit
we find
$$\sqrt{n} \left(\frac{n}{4 n - 3}\right)^{2 n}$$
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
$$a_{n} = \sqrt{n} \left(\frac{n}{4 n - 3}\right)^{2 n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt{n} \left(\frac{n + 1}{4 n + 1}\right)^{- 2 n - 2} \left|{\left(\frac{n}{4 n - 3}\right)^{2 n}}\right|}{\sqrt{n + 1}}\right)$$
Let's take the limit
we find
False
False
The rate of convergence of the power series
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