Represent $\frac1{(1+x)^2}$ as a power series

Question

Represent $\frac1{(1+x)^2}$ as a power series.

if we put $\frac1{1+2x+x^2}$:

$$ \sum _{n=1}^{+\infty}\left(-x-x^2\right)^n $$

and if we differentiate $\frac{-1}{1+x}$:

$$ -\sum _{n=1}^{+\infty}(-1)^nn\cdot x^{n-1} $$

are the expressions right?

thanks!

Answer 1

The geometric sum would be:

$\displaystyle \sum_{n=0}^\infty \left(-2x-x^2\right)^n$

Geometric sums start the summation at $n=0$. Also, your denominator is $1-(-2x-x^2)$, not $1-(-x-x^2)$.

The power series you wrote for the derivative of $-\dfrac{1}{1+x}$ is correct.

Answer 2

Represent $$\frac{1}{(x+1)^2}$$ as an infinite series.

Use the geometric sequence to get,

$$\sum_{n=0}^{\infty} x^n=\frac {1}{1-x}$$, replace $x$ by $-x$ $$\sum_{n=0}^{\infty} (-x)^n=\frac {1}{1-(-x)}=\frac {1}{1+x}$$ take the derivative, $$\sum_{n=1}^{\infty} n(-1)^n(x)^{n-1}=\frac {-1}{(1+x)^2}$$ Multiplying by $-1$ we get $$\sum_{n=1}^{\infty} n(-x)^{n-1}=\frac {1}{(1+x)^2}$$.

To answer your question, Yes.

Answer 3

Here is particularly simple method:

Just divide 1 by $(1 + x)^2$

To do this we expand $(1 + x)^2 = 1 + 2x + x^2$

Now we divide the one by this quantity

$$\frac{1}{1 + 2x + x^2} = 1 - \frac{2x + x^2}{1 + 2x + x^2} $$

And perform the next phase of division

$$ 1 - \frac{2x + x^2}{1 + 2x + x^2} = 1 - 2x + \frac{ x^2 + x^3}{1 + 2x + x^2}$$

And keep repeating:

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

Which will generate all the terms you seek. Now observe that on each round of division the leading term on the expression (the lower power one) being divided ends up in the series and the term right after that (always having coefficient one less than the coefficient before it) will be added (or subtracted) with 2 times the previous coefficient meaning it will have coefficient $(k-1) + 2(k+1) = k -1 + 2k + 2 = k +1$ meaning every coefficient will have magnitude one greater than the one preceding it.

That allows you to confidently say:

$$ \sum_{i=0}^\infty{(-1)^i(i+1)x^i} = \frac{1}{(1+x)^2} $$