Mister Exam
Other calculators
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- Product of series
-
How to use it?
- Sum of series:
-
(3-2i-5i^2+8i^3)
-
ln(k+1/k+2)
- 1/(2k+1)^2
- (ln(n)/n)^p
-
Identical expressions
- (three - two i-5i^2+8i^ three)
- (3 minus 2i minus 5i squared plus 8i cubed )
- (three minus two i minus 5i squared plus 8i to the power of three)
- (3-2i-5i2+8i3)
- 3-2i-5i2+8i3
- (3-2i-5i²+8i³)
- (3-2i-5i to the power of 2+8i to the power of 3)
- 3-2i-5i^2+8i^3
-
Similar expressions
- (3+2i-5i^2+8i^3)
- (3-2i+5i^2+8i^3)
- (3-2i-5i^2-8i^3)
Sum of series (3-2i-5i^2+8i^3)
The solution
You have entered
[src]
oo ___ \ ` \ / 2 3\ / \3 - 2*i - 5*i + 8*i / /__, i = 1
$$\sum_{i=1}^{\infty} \left(8 i^{3} + \left(- 5 i^{2} + \left(3 - 2 i\right)\right)\right)$$
Sum(3 - 2*i - 5*i^2 + 8*i^3, (i, 1, oo))
The radius of convergence of the power series
Given number:
$$8 i^{3} + \left(- 5 i^{2} + \left(3 - 2 i\right)\right)$$
It is a series of species
$$a_{i} \left(c x - x_{0}\right)^{d i}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{i \to \infty} \left|{\frac{a_{i}}{a_{i + 1}}}\right|}{c}$$
In this case
$$a_{i} = 8 i^{3} - 5 i^{2} - 2 i + 3$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{i \to \infty}\left(\frac{\left|{8 i^{3} - 5 i^{2} - 2 i + 3}\right|}{- 2 i + 8 \left(i + 1\right)^{3} - 5 \left(i + 1\right)^{2} + 1}\right)$$
Let's take the limit
we find
$$8 i^{3} + \left(- 5 i^{2} + \left(3 - 2 i\right)\right)$$
It is a series of species
$$a_{i} \left(c x - x_{0}\right)^{d i}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{i \to \infty} \left|{\frac{a_{i}}{a_{i + 1}}}\right|}{c}$$
In this case
$$a_{i} = 8 i^{3} - 5 i^{2} - 2 i + 3$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{i \to \infty}\left(\frac{\left|{8 i^{3} - 5 i^{2} - 2 i + 3}\right|}{- 2 i + 8 \left(i + 1\right)^{3} - 5 \left(i + 1\right)^{2} + 1}\right)$$
Let's take the limit
we find
True
False
The rate of convergence of the power series
The answer
[src]
oo ___ \ ` \ / 2 3\ / \3 - 5*i - 2*i + 8*i / /__, i = 1
$$\sum_{i=1}^{\infty} \left(8 i^{3} - 5 i^{2} - 2 i + 3\right)$$
Sum(3 - 5*i^2 - 2*i + 8*i^3, (i, 1, oo))
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