**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.

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}}\;=\;SIf 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.