Mister Exam
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How to use it?
- Sum of series:
- nx^n/(8n-9)*3^n
-
sqrt(n+1)-sqrt(n)/sqrt(n^2+n)
-
cos(i*n)/2^n
-
(3*x^2-2*x+3)/factorial(x+1)
-
Identical expressions
- one /ln(five ^n)
- 1 divide by ln(5 to the power of n)
- one divide by ln(five to the power of n)
- 1/ln(5n)
- 1/ln5n
- 1/ln5^n
- 1 divide by ln(5^n)
Sum of series 1/ln(5^n)
The solution
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oo ____ \ ` \ 1 \ ------- / / n\ / log\5 / /___, n = 1
$$\sum_{n=1}^{\infty} \frac{1}{\log{\left(5^{n} \right)}}$$
Sum(1/log(5^n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{\log{\left(5^{n} \right)}}$$
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
$$a_{n} = \frac{1}{\log{\left(5^{n} \right)}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\log{\left(5^{n + 1} \right)}}{\log{\left(5^{n} \right)}}}\right|$$
Let's take the limit
we find
$$\frac{1}{\log{\left(5^{n} \right)}}$$
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
$$a_{n} = \frac{1}{\log{\left(5^{n} \right)}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\log{\left(5^{n + 1} \right)}}{\log{\left(5^{n} \right)}}}\right|$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
Numerical answer
The series diverges
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