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:
- n^2*x^n/4^n
-
3/(8^(n-1))
- (3x)^2n/ln(2n-1)
-
(16-16cos(3(pi)n/4)/(pi^2n^2))cos(n(pi))
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Identical expressions
- (3x)^2n/ln(2n- one)
- (3x) squared n divide by ln(2n minus 1)
- (3x) squared n divide by ln(2n minus one)
- (3x)2n/ln(2n-1)
- 3x2n/ln2n-1
- (3x)²n/ln(2n-1)
- (3x) to the power of 2n/ln(2n-1)
- 3x^2n/ln2n-1
- (3x)^2n divide by ln(2n-1)
-
Similar expressions
- (3x)^2n/ln(2n+1)
Sum of series (3x)^2n/ln(2n-1)
The solution
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[src]
oo ____ \ ` \ 2 \ (3*x) *n / ------------ / log(2*n - 1) /___, n = 2
$$\sum_{n=2}^{\infty} \frac{n \left(3 x\right)^{2}}{\log{\left(2 n - 1 \right)}}$$
Sum(((3*x)^2*n)/log(2*n - 1), (n, 2, oo))
The radius of convergence of the power series
Given number:
$$\frac{n \left(3 x\right)^{2}}{\log{\left(2 n - 1 \right)}}$$
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
$$a_{n} = \frac{9 n x^{2}}{\log{\left(2 n - 1 \right)}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{n \log{\left(2 n + 1 \right)} \left|{\frac{1}{\log{\left(2 n - 1 \right)}}}\right|}{n + 1}\right)$$
Let's take the limit
we find
$$\frac{n \left(3 x\right)^{2}}{\log{\left(2 n - 1 \right)}}$$
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
$$a_{n} = \frac{9 n x^{2}}{\log{\left(2 n - 1 \right)}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{n \log{\left(2 n + 1 \right)} \left|{\frac{1}{\log{\left(2 n - 1 \right)}}}\right|}{n + 1}\right)$$
Let's take the limit
we find
True
False
The answer
[src]
oo ____ \ ` \ 2 \ 9*n*x / ------------- / log(-1 + 2*n) /___, n = 2
$$\sum_{n=2}^{\infty} \frac{9 n x^{2}}{\log{\left(2 n - 1 \right)}}$$
Sum(9*n*x^2/log(-1 + 2*n), (n, 2, oo))
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