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:
- x^n/n
-
1/(4n^2-1)
-
log((n+2)/n)^((-1)^(n+1))
- a^n*cos(a)+b^n*sin(n*b)
-
Identical expressions
- one /((4n- three)*(4n+ one))
- 1 divide by ((4n minus 3) multiply by (4n plus 1))
- one divide by ((4n minus three) multiply by (4n plus one))
- 1/((4n-3)(4n+1))
- 1/4n-34n+1
- 1 divide by ((4n-3)*(4n+1))
-
Similar expressions
- 1/((4n+3)*(4n+1))
- 1/(4*n-3)*(4*n+1)
- 1/((4*n-3)(4n+1))
- 1/(4n-3)(4n+1)
- 1/(4*n-3)(4*n+1)
- 1/((4n-3)*(4n-1))
Sum of series 1/((4n-3)*(4n+1))
The solution
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[src]
oo ___ \ ` \ 1 ) ------------------- / (4*n - 3)*(4*n + 1) /__, n = 1
$$\sum_{n=1}^{\infty} \frac{1}{\left(4 n - 3\right) \left(4 n + 1\right)}$$
Sum(1/((4*n - 3)*(4*n + 1)), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{\left(4 n - 3\right) \left(4 n + 1\right)}$$
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
$$a_{n} = \frac{1}{\left(4 n - 3\right) \left(4 n + 1\right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(4 n + 5\right) \left|{\frac{1}{4 n - 3}}\right|\right)$$
Let's take the limit
we find
$$\frac{1}{\left(4 n - 3\right) \left(4 n + 1\right)}$$
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
$$a_{n} = \frac{1}{\left(4 n - 3\right) \left(4 n + 1\right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(4 n + 5\right) \left|{\frac{1}{4 n - 3}}\right|\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
The answer
[src]
Gamma(9/4) ------------ 5*Gamma(5/4)
$$\frac{\Gamma\left(\frac{9}{4}\right)}{5 \Gamma\left(\frac{5}{4}\right)}$$
gamma(9/4)/(5*gamma(5/4))
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