# `f(x)=3/(3x+4) ,c=0` Find a power series for the function, centered at c and determine the interval of convergence.

A power series centered at `c=0` is follows the formula:

`sum_(n=0)^oo a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+...`

The given function `f(x)= 3/(3x+4)` resembles the power series:

`(1+x)^k = sum_(n=0)^oo (k(k-1)(k-2)...(k-n+1))/(n!) x ^n`

or

` (1+x)^k = 1+kx +(k(k-1))/(2!)x^2+(k(k-1)(k-2))/(3!)x^3+(k(k-1)(k-2)(k-3))/(4!)x^4+...`

For better comparison, we let `3x+4 = 4 ((3x)/4 + 1)` . The function becomes:

`f(x)= 3/4 ((3x)/4 + 1)`

Apply Law of exponents: `1/x^n = x^(-n)` .

`f(x)= 3/4((3x)/4 + 1)^(-1)`

Apply the aforementioned formula for power series on  `((3x)/4 + 1)^(-1)` , we may replace "x" with "`(3x)/4` " and "`k` " with "`-1` ". We let:

`(1+(3x)/4)^(-1) = sum_(n=0)^oo (-1(-1-1)(-1-2)...(-1-n+1))/(n!) ((3x)/4) ^n`

`=sum_(n=0)^oo (-1(-2)(-3)...(-1-n+1))/(n!)((3x)/4) ^n`

`=1+(-1)((3x)/4) +(-1(-2))/(2!)((3x)/4)^2+(-1(-2)(-3))/(3!)((3x)/4)^3+(-1(-2)(-3)(-4)/(4!)((3x)/4)^4+...`

`=1-(3x)/4 +(2)/2((3x)/4)^2- 6/6((3x)/4)^3+24/24((3x)/4)^4+...`

`=1-(3x)/4 +((3x)/4)^2- ((3x)/4)^3+((3x)/4)^4+...`

`=1-(3x)/4 +(9x^2)/16- (27x^3)/64+(81x^4)/256+...`

Applying `(1+(3x)/4)^(-1) =1-(3x)/4 +(9x^2)/16- (27x^3)/64+(81x^4)/256+...`  we get:

`3/4((3x)/4 + 1)^(-1)= 3/4*[1-(3x)/4 +(9x^2)/16- (27x^3)/64+(81x^4)/256+...]`

`=3/4-(9x)/16 +(27x^2)/64- (81x^3)/256+(243x^4)/1024+...`

`= sum_(n=0)^oo (-1)^n(3/4)^(n+1)x^n`

The power series of the function `f(x)=3/(3x+4)` centered at `c=0` is:

`3/(3x+4)=sum_(n=0)^oo (-1)^n(3/4)^(n+1)x^n`

or

`3/(3x+4)=3/4-(9x)/16 +(279x^2)/64- (81x^3)/256+(243x^4)/1024+...`

To determine the interval of convergence, we may apply geometric series test wherein the series `sum_(n=0)^oo a*r^n`  is convergent if `|r|lt1`  or `-1 ltrlt 1` . If `|r|gt=1` then the geometric series diverges.

Applying `(3/4)^(n+1) = (3/4)^n * (3/4)` on the series `sum_(n=0)^oo (-1)^n(3/4)^(n+1)x^n` , we get:

`sum_(n=0)^oo (-1)^n(3/4)^n(3/4)x^n =sum_(n=0)^oo(3/4) (-(3x)/4)^n`

By comparing `sum_(n=0)^oo(3/4) (-(3x)/4)^n` with  `sum_(n=0)^oo a*r^n` , we determine:`r =-(3x)/4` .

Apply the condition for convergence of geometric series: `|r|lt1` .

`|-(3x)/4|lt1`

`|-1| *|(3x)/4|lt1`

`1 *|(3x)/4|lt1`

`|(3x)/4|lt1`

`-1lt(3x)/4lt1`

Multiply each sides by `4/3` :

`-1*4/3lt(3x)/4*4/3lt1*4/3`

`-4/3 ltxlt4/3`

Check the convergence at endpoints that may satisfy `|(3x)/4|=1` .

Let `x=-4/3` on `sum_(n=0)^oo(3/4) (-(3x)/4)^n` , we get:

`sum_(n=0)^oo(3/4) (-3/4*-4/3)^n=sum_(n=0)^oo(1)^n`

Using geometric series test,  the ` r =1` satisfy `|r| gt=1` . Thus, the series diverges at `x=-4/3` .

Let `x=4/3` on `sum_(n=0)^oo(3/4) (-(3x)/4)^n` , we get:

`sum_(n=0)^oo(3/4) (-3/4*4/3)^n=sum_(n=0)^oo(-1)^n`

Using geometric series test,  the `r =-1` satisfy `|r| gt=1` . Thus, the series diverges at `x=-4/3` .

Thus, the power series `sum_(n=0)^oo (-1)^n(3/4)^(n+1)x^n` has an interval of convergence: `-4/3 ltxlt4/3` .

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