# power series

Power Series:

Power series is in the form of:

Σcnx^n=c0+c1(x-a)+c2(x-a)^2+c3(x-a)3+………..

Where cn is a function of the nth term. Each coefficient eg.c1, c2 is the result when the number for the nth term is substituted into the cn formula.

Suppose cn=n!(n+1), cn for the 5th term, (c5) would be 5!(5+1).

Also, the power series is centred about a. The power series always converges at x=a.

When given a problem about power series in math tests, common questions include finding the radius of convergence and the interval of convergence.

Vocab:
**Radius of convergence: if a series converges when abs(x-a)<R, the radius of convergence is R**

## Example Question 1:Find the radius of convergence of Σ(x-3)n/n

Perform the ratio test on the function. Ratio test starts by taking the limit as n approaches infinity of the absolute value of an+1/an where a is the function of n or (n+1). Let an be (x-3)n /n

Lim as n----∞ abs(an+1/an)=

Lim as n---∞ of abs(\(\frac{(x-3)^(n+1)}{n+1}\)*\(\frac{n}{(x-3)^n}\))=

Lim as n----∞ of abs(\(\frac{(x-3)*n}{n+1}\))

Then you divide the numerator and denominator by n Lim as n----∞ of \(\frac{abs(x-3)}{1+(1/n)}\)-abs(x-3) as n-->infinity

The ratio test tells that the series converges if the limit as n---∞ approaches a number less than 1 and diverges if the limit as n----∞ approaches a number greater than 1.

The series converges when abs(x-3)<1 and diverges when abs(x-3)>1, so the radius of convergence is 1.

Note: suppose 2*abs(x-3)<1, and abs(x-3)<2, then the radius of convergence would be 2.

*Interval of convergence: All values of x which makes the series converge. *Test to see the interval of convergence: If abs(x-3)<1, -1<x-3<1, and 2<x<4.

To see if the series converges when x=2, and x=4:

Put 2 instead of x in the original series, Σ(x-3)n/n, and you get Σ(-1)n/n, which converges n approaches infinity.

Put 4 instead of x in the original series, Σ(x-3)n/n, and you get Σ1/n, which diverges as n approaches infinity.

So, the series is convergent when 2<=x<4.

Example 2: Find the radius of convergence of Σn!(xn)

Perform the ratio test on the function.

Let an be n!(x^n)

Lim as n----∞ of abs(\(\frac{an+1}{an}\))=

Lim as n---∞ of abs(\(\frac{(n+1)!x^(n+1)}{n!x^n}\))=

Lim as n----∞ of (n+1)abs(x)=infinity

This series diverges when x does not equal 0. When the limit approaches infinity, the series only converges when x=0, and thus the radius of convergence is 0.

Example 3: Find the radius of convergence of Σ(-1)^n(x^2n/)(2^(2n)*(n!)2)Perform the ratio test on the function.

Let an be (-1)^n(x^2n/)(2^(2n)*(n!)2) then Lim as n----∞ abs(an+1/an)=

Lim as n----∞ of Abs(\(\frac{(-1)^(n+1)x^2(n+1)}{2^2(n+1)[(n+1)!]^2}\)*\(\frac{2^2n*(n!)^2}{(-1)^n*x^2n}\))=

Lim as n---∞ of abs(\(\frac{x^(2n+2)}{2^(2n+2)*(n+1)^2(n!)^2}\)*\(\frac{2^2n*(n!)^2}{x^2n}\)) = \(\frac{x^2}{4(n+1)^2}\)-->0<1 for all x

When the limit approaches zero, the series converges when x is any real number. Thus, the radius of convergence is infinity, and the interval of convergence is from negative infinity to infinity.

Examples 1 and 2 taken from Calculus 7E Early Transcendentals by James Stewart.

Special Thanks to Thomas Shelly, math professor at Mount Holyoke College