Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
- Differential equations Step by Step
- Product of series
-
How to use it?
- Sum of series:
-
(3x^2-2x+3)(x+1)^x
-
sqrt(n)*(n/(4*n-3))^(2*n)
-
arcctg((n+1)/(n^2-3))
-
exp(-0.01*n)
-
Identical expressions
- two /(ten ^(4n+ one))
- 2 divide by (10 to the power of (4n plus 1))
- two divide by (ten to the power of (4n plus one))
- 2/(10(4n+1))
- 2/104n+1
- 2/10^4n+1
- 2 divide by (10^(4n+1))
-
Similar expressions
- 2/(10^(4n-1))
Sum of series 2/(10^(4n+1))
The solution
You have entered
[src]
oo ____ \ ` \ 2 \ --------- / 4*n + 1 / 10 /___, n = 1
$$\sum_{n=1}^{\infty} \frac{2}{10^{4 n + 1}}$$
Sum(2/10^(4*n + 1), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{2}{10^{4 n + 1}}$$
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
$$a_{n} = 2 \cdot 10^{- 4 n - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(10^{- 4 n - 1} \cdot 10^{4 n + 5}\right)$$
Let's take the limit
we find
$$\frac{2}{10^{4 n + 1}}$$
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
$$a_{n} = 2 \cdot 10^{- 4 n - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(10^{- 4 n - 1} \cdot 10^{4 n + 5}\right)$$
Let's take the limit
we find
False
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