Mister Exam
Other calculators
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- Fourier series step by step
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- Product of series
-
How to use it?
- Sum of series:
-
cos(2n)/n^2
- factorial(n)^2/factorial(k^n)
-
3^(n-1)/7^n
-
cos(npi*2)/n^4
-
Identical expressions
- three ^(n- one)/ seven ^n
- 3 to the power of (n minus 1) divide by 7 to the power of n
- three to the power of (n minus one) divide by seven to the power of n
- 3(n-1)/7n
- 3n-1/7n
- 3^n-1/7^n
- 3^(n-1) divide by 7^n
-
Similar expressions
- 3^(n+1)/7^n
Sum of series 3^(n-1)/7^n
The solution
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oo ____ \ ` \ n - 1 \ 3 ) ------ / n / 7 /___, n = 1
$$\sum_{n=1}^{\infty} \frac{3^{n - 1}}{7^{n}}$$
Sum(3^(n - 1)/7^n, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{3^{n - 1}}{7^{n}}$$
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
$$a_{n} = 3^{n - 1}$$
and
$$x_{0} = -7$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-7 + \lim_{n \to \infty}\left(3^{- n} 3^{n - 1}\right)\right)$$
Let's take the limit
we find
$$R = 0$$
$$\frac{3^{n - 1}}{7^{n}}$$
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
$$a_{n} = 3^{n - 1}$$
and
$$x_{0} = -7$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-7 + \lim_{n \to \infty}\left(3^{- n} 3^{n - 1}\right)\right)$$
Let's take the limit
we find
False
False
$$R = 0$$
The rate of convergence of the power series
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