Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
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- Product of series
-
How to use it?
- Sum of series:
- ((2*n+4)/(5*n+7))^n*x^n
-
log((n+2)/n)^((-1)^(n+1))
- x^n*4^n/((5^n*(n+9)^(1/9)))
-
4^n-2*25^n+1/5^4n-2
-
Identical expressions
- four ^n- two * twenty-five ^n+ one / five ^4n- two
- 4 to the power of n minus 2 multiply by 25 to the power of n plus 1 divide by 5 to the power of 4n minus 2
- four to the power of n minus two multiply by twenty minus five to the power of n plus one divide by five to the power of 4n minus two
- 4n-2*25n+1/54n-2
- 4^n-2*25^n+1/5⁴n-2
- 4^n-225^n+1/5^4n-2
- 4n-225n+1/54n-2
- 4^n-2*25^n+1 divide by 5^4n-2
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Similar expressions
- 4^n-2*25^n+1/5^4n+2
- 4^n-2*25^n-1/5^4n-2
- 4^n+2*25^n+1/5^4n-2
Sum of series 4^n-2*25^n+1/5^4n-2
The solution
You have entered
[src]
oo ____ \ ` \ / n n 1 \ \ |4 - 2*25 + --*n - 2| / | 4 | / \ 5 / /___, n = 0
$$\sum_{n=0}^{\infty} \left(\left(\frac{n}{625} + \left(- 2 \cdot 25^{n} + 4^{n}\right)\right) - 2\right)$$
Sum(4^n - 2*25^n + (1/5)^4*n - 2, (n, 0, oo))
The radius of convergence of the power series
Given number:
$$\left(\frac{n}{625} + \left(- 2 \cdot 25^{n} + 4^{n}\right)\right) - 2$$
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
$$a_{n} = - 2 \cdot 25^{n} + 4^{n} + \frac{n}{625} - 2$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{2 \cdot 25^{n} - 4^{n} - \frac{n}{625} + 2}{2 \cdot 25^{n + 1} - 4^{n + 1} - \frac{n}{625} + \frac{1249}{625}}}\right|$$
Let's take the limit
we find
$$\left(\frac{n}{625} + \left(- 2 \cdot 25^{n} + 4^{n}\right)\right) - 2$$
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
$$a_{n} = - 2 \cdot 25^{n} + 4^{n} + \frac{n}{625} - 2$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{2 \cdot 25^{n} - 4^{n} - \frac{n}{625} + 2}{2 \cdot 25^{n + 1} - 4^{n + 1} - \frac{n}{625} + \frac{1249}{625}}}\right|$$
Let's take the limit
we find
False
False
False
The rate of convergence of the power series
The answer
[src]
oo ___ \ ` \ / n n n \ ) |-2 + 4 - 2*25 + ---| / \ 625/ /__, n = 0
$$\sum_{n=0}^{\infty} \left(- 2 \cdot 25^{n} + 4^{n} + \frac{n}{625} - 2\right)$$
Sum(-2 + 4^n - 2*25^n + n/625, (n, 0, oo))
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