Mister Exam
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-
How to use it?
- Sum of series:
-
1/(n(n+2))
-
(2n-1)/2^n
-
sin(3n)/(7n)^(1/5)
-
arctg1/(2n^2)
-
Identical expressions
- sin(3n)/(7n)^(one / five)
- sinus of (3n) divide by (7n) to the power of (1 divide by 5)
- sinus of (3n) divide by (7n) to the power of (one divide by five)
- sin(3n)/(7n)(1/5)
- sin3n/7n1/5
- sin3n/7n^1/5
- sin(3n) divide by (7n)^(1 divide by 5)
Sum of series sin(3n)/(7n)^(1/5)
The solution
You have entered
[src]
oo ____ \ ` \ sin(3*n) \ -------- / 5 _____ / \/ 7*n /___, n = 1
$$\sum_{n=1}^{\infty} \frac{\sin{\left(3 n \right)}}{\sqrt[5]{7 n}}$$
Sum(sin(3*n)/(7*n)^(1/5), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\sin{\left(3 n \right)}}{\sqrt[5]{7 n}}$$
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
$$a_{n} = \frac{7^{\frac{4}{5}} \sin{\left(3 n \right)}}{7 \sqrt[5]{n}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt[5]{n + 1} \left|{\frac{\sin{\left(3 n \right)}}{\sin{\left(3 n + 3 \right)}}}\right|}{\sqrt[5]{n}}\right)$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt[5]{n + 1} \left|{\frac{\sin{\left(3 n \right)}}{\sin{\left(3 n + 3 \right)}}}\right|}{\sqrt[5]{n}}\right)$$
$$\frac{\sin{\left(3 n \right)}}{\sqrt[5]{7 n}}$$
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
$$a_{n} = \frac{7^{\frac{4}{5}} \sin{\left(3 n \right)}}{7 \sqrt[5]{n}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt[5]{n + 1} \left|{\frac{\sin{\left(3 n \right)}}{\sin{\left(3 n + 3 \right)}}}\right|}{\sqrt[5]{n}}\right)$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt[5]{n + 1} \left|{\frac{\sin{\left(3 n \right)}}{\sin{\left(3 n + 3 \right)}}}\right|}{\sqrt[5]{n}}\right)$$
False
The rate of convergence of the power series
The answer
[src]
oo ____ \ ` \ 4/5 \ 7 *sin(3*n) ) ------------- / 5 ___ / 7*\/ n /___, n = 1
$$\sum_{n=1}^{\infty} \frac{7^{\frac{4}{5}} \sin{\left(3 n \right)}}{7 \sqrt[5]{n}}$$
Sum(7^(4/5)*sin(3*n)/(7*n^(1/5)), (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