Mister Exam
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- Taylor Series Step by Step
- Fourier series step by step
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- Product of series
-
How to use it?
- Sum of series:
-
sqrt((n+4)/((n^4)+4))
-
1/((3*n+1)*(3*n+4))
-
lnn/n
-
sin(pi/2^(n-1))
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Identical expressions
- eight /(16n^ two +8n- fifteen)
- 8 divide by (16n squared plus 8n minus 15)
- eight divide by (16n to the power of two plus 8n minus fifteen)
- 8/(16n2+8n-15)
- 8/16n2+8n-15
- 8/(16n²+8n-15)
- 8/(16n to the power of 2+8n-15)
- 8/16n^2+8n-15
- 8 divide by (16n^2+8n-15)
-
Similar expressions
- 8/(16n^2+8n+15)
- 8/(16n^2-8n-15)
Sum of series 8/(16n^2+8n-15)
The solution
You have entered
[src]
oo ____ \ ` \ 8 \ ---------------- / 2 / 16*n + 8*n - 15 /___, n = 1
$$\sum_{n=1}^{\infty} \frac{8}{\left(16 n^{2} + 8 n\right) - 15}$$
Sum(8/(16*n^2 + 8*n - 15), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{8}{\left(16 n^{2} + 8 n\right) - 15}$$
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
$$a_{n} = \frac{8}{16 n^{2} + 8 n - 15}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(8 n + 16 \left(n + 1\right)^{2} - 7\right) \left|{\frac{1}{16 n^{2} + 8 n - 15}}\right|\right)$$
Let's take the limit
we find
$$\frac{8}{\left(16 n^{2} + 8 n\right) - 15}$$
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
$$a_{n} = \frac{8}{16 n^{2} + 8 n - 15}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\left(8 n + 16 \left(n + 1\right)^{2} - 7\right) \left|{\frac{1}{16 n^{2} + 8 n - 15}}\right|\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
The answer
[src]
8*Gamma(13/4) ------------- 15*Gamma(9/4)
$$\frac{8 \Gamma\left(\frac{13}{4}\right)}{15 \Gamma\left(\frac{9}{4}\right)}$$
8*gamma(13/4)/(15*gamma(9/4))
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