Question

How to find the Maclaurin series and the radius of convergence for `f(x)=1/(1+x)^2`?

Answer 1

The Maclaurin series is given by `sum_(n= 1)^oo (-1)^(n+ 1)(n)x^(n - 1)` and the radius of convergence is `1`

Explanation:

Recall that the McLaurin series is given by

`f(0) + (f'(0)x)/(1!) + (f''(0) x^2)/(2!) +... + (f^n(0) x^n)/(n!)`

Let's start by finding the value of `f(0)` on the first few derivatives.

`f(0) = 1/(1 + 0)^2 = 1`

`f'(x) = (-2(x + 1))/(x + 1)^4 = -2/(x + 1)^3`
`f'(0) = -2/(0 +1)^3 = -2`

`f''(x) = 6(x + 1)^2/(x + 1)^6 = 6/(x + 1)^4`
`f''(0) = 6`

Now let's recall the first few factorials.

`1! = 1`
`2! = 2`
`3! = 6`
`...`

We can see the similarity here. Our first few terms are therefore

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

Which can be written as `sum_(n= 1)^oo (-1)^(n+ 1)(n)x^(n - 1)`

The radius of converge is given by the ratio test.

`=lim_(n -> oo) ((-1)^(n + 1+ 1)(n +1)x^(n + 1- 1))/((-1)^(n +1)n(x^(n - 1))`

`=lim_(n -> oo) ((-1)^(n + 2)(n + 1)x^n)/((-1)^(n + 1)(n)x^(n - 1))`

`=lim_(n-> oo) (-1)^( 1)(n + 1)/n x`

`=(-1)(1)x`

`=-x`

By the ratio test, `|x| < 1` for the series to be convergent. Therefore, the radius of convergence is `1`.

Hopefully this helps!