Differentiating a power series

[tmt1]

I need to prove that for $-1 < x < 1$

$$\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + 4x^3 ...$$

So, according to the textbook, the geometric series has a radius of convergence $R = 1$ (I'm not sure how this is true).

In any case we can compare it to:

$$\frac{1}{1 - x} =\sum_{n = 0}^{\infty} x^n$$

If we differentiate it, we will get:

$$1 + \sum_{n = 1}^{\infty} (n + 1) x^n$$

(Or, $1 + 2x + 3x^2 + 4x^3 + 5x^4 ...$)

So, somehow I need to prove that $\frac{1}{(1 - x)^2}$ is equal to the derivative of $\sum_{n = 0}^{\infty} x^n$ and I'm not sure how to do that.

 

[I like Serena]

I need to prove that for $-1 < x < 1$

$$\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + 4x^3 ...$$

So, according to the textbook, the geometric series has a radius of convergence $R = 1$ (I'm not sure how this is true).

In any case we can compare it to:

$$\frac{1}{1 - x} =\sum_{n = 0}^{\infty} x^n$$

If we differentiate it, we will get:

$$1 + \sum_{n = 1}^{\infty} (n + 1) x^n$$

(Or, $1 + 2x + 3x^2 + 4x^3 + 5x^4 ...$)

So, somehow I need to prove that $\frac{1}{(1 - x)^2}$ is equal to the derivative of $\sum_{n = 0}^{\infty} x^n$ and I'm not sure how to do that.

Hi tmt! ;)

You have found that if you take the derivative of $\sum x^n$, you get a match for the series.
What do you get if you take the derivative of $\frac 1{1-x}$, which is equal to that series?