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:
-
(4*9^n-2^n)/18^n
-
1/((3*n+1)*(3*n+4))
- sqrt(n)*(x+3)^n/n^2+1
-
ln((n(n+2))/(n+1)^2)
-
Identical expressions
- [(- one)^(k+ one)*((four)/(three))^(k)+((four)/(five))^(k- one)]
- [( minus 1) to the power of (k plus 1) multiply by ((4) divide by (3)) to the power of (k) plus ((4) divide by (5)) to the power of (k minus 1)]
- [( minus one) to the power of (k plus one) multiply by ((four) divide by (three)) to the power of (k) plus ((four) divide by (five)) to the power of (k minus one)]
- [(-1)(k+1)*((4)/(3))(k)+((4)/(5))(k-1)]
- [-1k+1*4/3k+4/5k-1]
- [(-1)^(k+1)((4)/(3))^(k)+((4)/(5))^(k-1)]
- [(-1)(k+1)((4)/(3))(k)+((4)/(5))(k-1)]
- [-1k+14/3k+4/5k-1]
- [-1^k+14/3^k+4/5^k-1]
- [(-1)^(k+1)*((4) divide by (3))^(k)+((4) divide by (5))^(k-1)]
-
Similar expressions
- [(-1)^(k+1)*((4)/(3))^(k)+((4)/(5))^(k+1)]
- [(-1)^(k+1)*((4)/(3))^(k)-((4)/(5))^(k-1)]
- [(1)^(k+1)*((4)/(3))^(k)+((4)/(5))^(k-1)]
- [(-1)^(k-1)*((4)/(3))^(k)+((4)/(5))^(k-1)]
Sum of series [(-1)^(k+1)*((4)/(3))^(k)+((4)/(5))^(k-1)]
The solution
You have entered
[src]
oo ___ \ ` \ / k + 1 k k - 1\ / \(-1) *1.33333333333333 + 0.8 / /__, n = 5
$$\sum_{n=5}^{\infty} \left(\left(-1\right)^{k + 1} \cdot 1.33333333333333^{k} + 0.8^{k - 1}\right)$$
Sum((-1)^(k + 1)*1.33333333333333^k + 0.8^(k - 1), (n, 5, oo))
The radius of convergence of the power series
Given number:
$$\left(-1\right)^{k + 1} \cdot 1.33333333333333^{k} + 0.8^{k - 1}$$
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
$$a_{n} = \left(-1\right)^{k + 1} \cdot 1.33333333333333^{k} + 0.8^{k - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} 1$$
Let's take the limit
we find
$$\left(-1\right)^{k + 1} \cdot 1.33333333333333^{k} + 0.8^{k - 1}$$
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
$$a_{n} = \left(-1\right)^{k + 1} \cdot 1.33333333333333^{k} + 0.8^{k - 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} 1$$
Let's take the limit
we find
True
False
The answer
[src]
/ -1 + k 1 + k k\ oo*\0.8 + (-1) *1.33333333333333 /
$$\infty \left(\left(-1\right)^{k + 1} \cdot 1.33333333333333^{k} + 0.8^{k - 1}\right)$$
oo*(0.8^(-1 + k) + (-1)^(1 + k)*1.33333333333333^k)
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