Mister Exam
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How to use it?
- Sum of series:
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1/(n(n+2))
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(2n-1)/2^n
-
sqrt(n)-2*sqrt(n+1)+sqrt(n+2)
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log((n^2+1)/n^2)
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Identical expressions
- (i+ one)*(two ^(i- one))
- (i plus 1) multiply by (2 to the power of (i minus 1))
- (i plus one) multiply by (two to the power of (i minus one))
- (i+1)*(2(i-1))
- i+1*2i-1
- (i+1)(2^(i-1))
- (i+1)(2(i-1))
- i+12i-1
- i+12^i-1
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Similar expressions
- (i+1)*(2^(i+1))
- (i-1)*(2^(i-1))
Sum of series (i+1)*(2^(i-1))
The solution
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oo ___ \ ` \ i - 1 / (i + 1)*2 /__, i = 1
$$\sum_{i=1}^{\infty} 2^{i - 1} \left(i + 1\right)$$
Sum((i + 1)*2^(i - 1), (i, 1, oo))
The radius of convergence of the power series
Given number:
$$2^{i - 1} \left(i + 1\right)$$
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
$$a_{i} = 2^{i - 1} \left(i + 1\right)$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{i \to \infty}\left(\frac{2^{- i} 2^{i - 1} \left(i + 1\right)}{i + 2}\right)$$
Let's take the limit
we find
$$2^{i - 1} \left(i + 1\right)$$
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
$$a_{i} = 2^{i - 1} \left(i + 1\right)$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{i \to \infty}\left(\frac{2^{- i} 2^{i - 1} \left(i + 1\right)}{i + 2}\right)$$
Let's take the limit
we find
False
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