Mister Exam
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How to use it?
- Sum of series:
-
200(0.98)^n-1
-
sqrt(n)*(n/(4*n-3))^(2*n)
-
arcctg((n+1)/(n^2-3))
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3k
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Identical expressions
- two hundred (zero . ninety-eight)^n- one
- 200(0.98) to the power of n minus 1
- two hundred (zero . ninety minus eight) to the power of n minus one
- 200(0.98)n-1
- 2000.98n-1
- 2000.98^n-1
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Similar expressions
- 200(0.98)^n+1
Sum of series 200(0.98)^n-1
The solution
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oo ____ \ ` \ / n \ \ | /49\ | / |200*|--| - 1| / \ \50/ / /___, n = 1
$$\sum_{n=1}^{\infty} \left(200 \left(\frac{49}{50}\right)^{n} - 1\right)$$
Sum(200*(49/50)^n - 1, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$200 \left(\frac{49}{50}\right)^{n} - 1$$
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
$$a_{n} = 200 \left(\frac{49}{50}\right)^{n} - 1$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{200 \left(\frac{49}{50}\right)^{n} - 1}{200 \left(\frac{49}{50}\right)^{n + 1} - 1}}\right|$$
Let's take the limit
we find
$$200 \left(\frac{49}{50}\right)^{n} - 1$$
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
$$a_{n} = 200 \left(\frac{49}{50}\right)^{n} - 1$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{200 \left(\frac{49}{50}\right)^{n} - 1}{200 \left(\frac{49}{50}\right)^{n + 1} - 1}}\right|$$
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