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:
-
(4*9^n-2^n)/18^n
-
1/((3*n+1)*(3*n+4))
- sqrt(n)*(x+3)^n/n^2+1
-
ln((n(n+2))/(n+1)^2)
-
Identical expressions
- ln((n(n+ two))/(n+ one)^ two)
- ln((n(n plus 2)) divide by (n plus 1) squared )
- ln((n(n plus two)) divide by (n plus one) to the power of two)
- ln((n(n+2))/(n+1)2)
- lnnn+2/n+12
- ln((n(n+2))/(n+1)²)
- ln((n(n+2))/(n+1) to the power of 2)
- lnnn+2/n+1^2
- ln((n(n+2)) divide by (n+1)^2)
-
Similar expressions
- ln((n(n+2))/(n-1)^2)
- ln((n(n-2))/(n+1)^2)
Sum of series ln((n(n+2))/(n+1)^2)
The solution
You have entered
[src]
oo ____ \ ` \ /n*(n + 2)\ \ log|---------| / | 2| / \ (n + 1) / /___, n = 1
$$\sum_{n=1}^{\infty} \log{\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}} \right)}$$
Sum(log((n*(n + 2))/(n + 1)^2), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\log{\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}} \right)}$$
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
$$a_{n} = \log{\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}} \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\log{\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}} \right)}}{\log{\left(\frac{\left(n + 1\right) \left(n + 3\right)}{\left(n + 2\right)^{2}} \right)}}}\right|$$
Let's take the limit
we find
$$\log{\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}} \right)}$$
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
$$a_{n} = \log{\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}} \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\log{\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}} \right)}}{\log{\left(\frac{\left(n + 1\right) \left(n + 3\right)}{\left(n + 2\right)^{2}} \right)}}}\right|$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
The answer
[src]
oo ____ \ ` \ /n*(2 + n)\ \ log|---------| / | 2| / \ (1 + n) / /___, n = 1
$$\sum_{n=1}^{\infty} \log{\left(\frac{n \left(n + 2\right)}{\left(n + 1\right)^{2}} \right)}$$
Sum(log(n*(2 + n)/(1 + n)^2), (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