Mister Exam
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How to use it?
- Sum of series:
- x^(n-1)
-
1/(n*lnn)
- (lnx-1)/(ln(x+1)-1)
- ln^n(x)/n!
-
Identical expressions
- one /(n*lnn)
- 1 divide by (n multiply by lnn)
- one divide by (n multiply by lnn)
- 1/(nlnn)
- 1/nlnn
- 1 divide by (n*lnn)
-
Similar expressions
- 1/(nln(n)ln(ln(n)))
- 1/(n(lnn)((ln)(lnn))^2)
- 1/(n((lnn)^2))
- 1/(n((ln*n)^2))
- 1/(nlnn)
Sum of series 1/(n*lnn)
The solution
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oo ___ \ ` \ 1 ) -------- / n*log(n) /__, n = 2
$$\sum_{n=2}^{\infty} \frac{1}{n \log{\left(n \right)}}$$
Sum(1/(n*log(n)), (n, 2, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{n \log{\left(n \right)}}$$
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
$$a_{n} = \frac{1}{n \log{\left(n \right)}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(n + 1 \right)} \left|{\frac{1}{\log{\left(n \right)}}}\right|}{n}\right)$$
Let's take the limit
we find
$$\frac{1}{n \log{\left(n \right)}}$$
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
$$a_{n} = \frac{1}{n \log{\left(n \right)}}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(n + 1 \right)} \left|{\frac{1}{\log{\left(n \right)}}}\right|}{n}\right)$$
Let's take the limit
we find
True
False
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