Define Convergence of Power Series with all Test

Define Convergence of Power Series?

An infinite series of the form

c_0 + c_1x + c_2x^2 + … + c_{n\;}x^n + … = \overset\infty{\underset0{\sum c_n}}x^n = 0 (1)

Is called a power series in x. the coefficient c_n are real number and x is a real variable.

Define Convergence of Power Series with all Test

A series of the form

c_0 + c_1\left(x-a\right) + c_2\left(x-a\right)^2 + … +c_n\left(x-a\right)^n + … = c_n\left(x- a\right)^n (2)

is a power series in \left(x-a\right) where a is a real number. However the series (2) can always be reduced to a series of the form (1) by putting \left(x-a\right)\;=y . But here we shall study the power series of form (1).

What is Convergence of Power Series?

If a numerical value of X is substituted into (1) then it is a series of the constant terms.

Let \left\{S_n\right\} be the sequence of partial sum of the series \overset\infty{\underset1{\sum a_n}}. If the sequence \left\{S_n\right\} converges to the limit S, then the \overset\infty{\underset1{\sum a_n}} is said to converge and S is called the sum of the series. In this case we write

\overset\infty{\underset1{\sum a_n}}\;=\;S

If the sequence \left\{S_n\right\} diverges then the series \overset\infty{\underset1{\sum a_n}} is said to be divergent.

Convergence of Power Series test

There are some tests to check whether the series is converged is diverge

  • The Basic Comparison Test
  • The Limit Comparison Test
  • The Integral Test
  • The Ratio Test
  • Cauchy’s Root Test

Define Basic Comparison Test?

Let \overset\infty{\underset1{\sum a_n}} and \overset\infty{\underset1{\sum b_n}} be series of positive terms with a_{n\;}\;\leq\;b_n for each n=1,2,3…. then

  • If \overset\infty{\underset1{\sum b_n}} converges, then \overset\infty{\underset1{\sum a_n}} is converges.
  • If \overset\infty{\underset1{\sum a_n}} diverges then the \overset\infty{\underset1{\sum b_n}} is diverge.

What is Limit Comparison Test?

Let \overset\infty{\underset1{\sum a_n}} and \overset\infty{\underset1{\sum b_n}} be series of positive terms. Then

  • If \lim_{n\rightarrow\infty}\frac{a_n}{b_n}\;=\;L\;\neq0, then either both series converge or both diverge.
  • If \lim_{n\rightarrow\infty}\frac{a_n}{b_n}\;=\;0 and \overset\infty{\underset1{\sum b_n}} converge then \overset\infty{\underset1{\sum a_n}} is also converges.
  • If \lim_{n\rightarrow\infty}\frac{a_n}{b_n}\;=\;\infty and \overset\infty{\underset1{\sum b_n}} diverges then \overset\infty{\underset1{\sum a_n}} is also diverges.

What is Integral Test?

Let \overset\infty{\underset1{\sum a_n}} be a positive term series.if f is the continous and nonincreasing function on [ 1, \infty ) such that f (n) = a_n for all positive integer n, then

  • \overset\infty{\underset1{\sum a_n}} convergers if \int_1^\infty f\left(x\right)\operatorname dx converges.
  • \overset\infty{\underset1{\sum a_n}} diverge if \int_1^\infty f\left(x\right)\operatorname dx diverges.

What is Ratio Test?

let \overset\infty{\underset1{\sum a_n}} be a serie sof positive terms and suppose tha \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\;=\;L , where L is a non negative real number.

  • If L< 1, the series \overset\infty{\underset1{\sum a_n}} converges.
  • If L> 1 ,or infinity the series \overset\infty{\underset1{\sum a_n}} converges.
  • If L= 1 the test fails to determine converges or diverges of the series.

What is Cauchy’s Root Test?

Let \overset\infty{\underset1{\sum a_n}} be a series of positive terms and suppose that \;\lim_{n\rightarrow\infty}\left(a_n\right)^{1/n}\;=L where L is a nonnegative real number or infinity.

  • If L< 1, the series [katex]\overset\infty{\underset1{\sum a_n}}[/katex] converges.
  • If L> 1 ,or infinity the series \overset\infty{\underset1{\sum a_n}} converges.
  • If L=1 the test fails.

Find the Interval of Convergence of the Power Series

The set of all values of X for which a power series converges  is called the interval of convergences

Define Radius of Convergence of Power Series

The number R in any interval is called the radius of convergence of the series.

How do you find the Radius of Convergence of Series?

Every power series has an interval of convergence and radius of convergence by the power series \sum_{n=0}^\infty\;c_nx^n converge the only for x= 0 then its interval of convergence is reduced to the point 0 and its radius of convergence is 0.

If the power series for all values of x, then its interval of convergence Is interval \left(-\infty\;\infty\right) and its the radius of convergence is Infinity.

If the power series converges for \left|x\right|\;<\;R[/katex] and diverges for [katex]\left|x\right|\;>\;R then the radius of converges is R ad its interval is one of the interval \left(-R,\;R\right) or \left[-R,\;R\right].

Addition and Multiplication of Power Series with Equation

Let \sum_{n=0}^\infty\;a_nx^n and \sum_{n=0}^\infty\;b_nx^n be two power series with a common interval of convergence.then

  • \sum_{n=0}^\infty\;a_nx^n + \sum_{n=0}^\infty\;b_nx^n = \sum_{n=0}^\infty\left(a_n+x_n\right)x^n
  • (\sum_{n=0}^\infty\;a_nx^n) (\sum_{n=0}^\infty\;b_nx^n ) = \sum_{n=0}^\infty\;c_nx^n

Where c_{n\;=\;\;}a_0b_n+a_1b_{n-1}+…+a_{n-1}b_1+a_nb_0 power series may be added and multiplied together much the same way as a polynomial.

  A power series may also be divided by another power series in a manner similar to the division of polynomials.

The interval of convergence of each of the new power series obtain by the algebraical operation is a common interval of convergence of the two given power series.

Differentiation and Integration of Power series

Each power series \sum_{n=0}^\infty\;c_nx^n defines a function f where

f(x) = \sum_{n=0}^\infty\;c_nx^n

for each X in the interval of convergence of the power series. \sum_{n=0}^\infty\;c_nx^n is called a power series representation of f(x). The function f is continuous and differentiable. Taylor and Maclaurin’s formula is used to obtain a power series representation of a function having derivatives of any order on the same interval. It is known that the geometric series 1+x+x^2+… converges for -1 < x< 1 and its sum is[katex]\frac1{1-x}[/katex]. thus a power series representation of [katex]\frac1{1-x}[/katex] is the power series [katex]1+x+x^2+…[/katex] whose radius of convergence is 1.

Having obtained a function f representative by power series. A natural question arises whether their function can be differentiated and integrated. The answer is that the new power series can be obtained by term-by-term differentiation and integration of a given power series within its interval of convergence.

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