How to prove this Taylor expansion of $\frac{1}{(1+x)^2}=-1\times\displaystyle\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$? I came across this series of the Taylor Expansion-

$$\frac{1}{(1+x)^2}=1 - 2x + 3x^2 -4x^3 + \dots.=-1\times\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$$

But I have no idea how to prove this...
Thanks for any help!
 A: Let $f(x) = 1/(1+x)$ and notice $f'(x) = -1/(1+x)^2$. However, $f(x) = 1-x+x^2-x^3+ \cdots$ hence $-f'(x) = 1-2x+3x^2-4x^3+\cdots$.
A: Notice that, by Newton's generalized binomial expansion, $$1/(1+x)^2= \sum^\infty_{k=0}{-2\choose k}x^{k}$$ and $${-2 \choose k}=\frac{(-2)(-3)...(-1-k)}{k!}=(-1)^k\frac{(k+1)!}{k!}=(-1)^k(k+1).$$
A: Here are two approaches.
METHODOLOGY $1$:  Non-Calculus Based
Recall the expansion of $\frac{1}{1+x}$ is given by
$$\frac{1}{1+x}=\sum_{k=0}^{\infty}(-x)^k$$
for $|x|<1$, which can be obtained by summing the geometric progression.  
Then, we have
$$\begin{align}\sum_{n=0}^{\infty}(-1)^n(n+1)x^n&=\sum_{n=0}^{\infty}(-1)^n x^n \sum_{k=0}^{n}(1)\\\\
&=\sum_{k=0}^\infty \sum_{n=k}^\infty (-x)^n\\\\
&=\sum_{k=0}^{\infty}\frac{(-x)^k}{1+x}\\\\
&=\frac{1}{(1+x)^2}
\end{align}$$
as was to be shown!

METHODOLOGY $2$:  Calculus Based
Recall that the expansion of $\frac{1}{1+x}$ is given by 
$$\frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n$$
for $|x|<1$, which can be obtained by summing the geometric progression.
Now, noting that $\frac{1}{(1+x)^2}=-\frac{d}{dx}\left(\frac{1}{1+x}\right)$, we have
$$\begin{align}
\frac{1}{(1+x)^2}&=-\frac{d}{dx}\left(\frac{1}{1+x}\right)\\\\
&=-\frac{d}{dx}\left(\sum_{n=0}^\infty (-1)^nx^n\right)\\\\
&=\sum_{n=0}^\infty (-1)^{n}\,(n+1) x^{n}
\end{align}$$  
as expected!
A: Observe that 
$$\frac x{(1+x)^2}=x - 2x^2 + 3x^3 -4x^4 + \dots$$ and adding the original
$$\frac1{(1+x)^2}=1-2x+3x^2-4x^3+5x^4\cdots$$
you verify
$$\frac x{(1+x)^2}+\frac1{(1+x)^2}=1-x+x^2-x^3+x^4-\cdots=\frac1{x+1}.$$
A: Expanding
$$(x^2+2x+1)(1 - 2x + 3x^2 -4x^3 + 5x^4-\dots),$$
the independent term is $1$, the linear term is $2x-2x$, then all next powers of $x$ get a coefficient which is the linear combination $1,2,1$ of three successive coefficients of the series, i.e. $0$.
A: Here isn't a VERFICATION of hte fact, but a full derivation, as if you had to come up with this question all by yourself to ask someone else.
Recall that taylor's theorem states:
Given an smooth function $f(x)$ around a point $a$ (that is the function is infinitely differentiable around a, so $f(a)$, $f'(a)$, $f''(a)$ all exist. 
Then it is the case that around this point
$$ f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \frac{1}{3!}f'''(a)(x-a)^3 + ... $$
So given:
$$ f(x) = \frac{1}{(1+x)^2}$$ 
Observe that 
$$ f'(x) = -\frac{2}{(1+x)^3} $$
$$ f''(x) = \frac{2 \times 3}{(1+x)^4}$$
$$ f'''(x) = -\frac{2 \times 3 \times 4}{(1+x)^4}$$
And catching a pattern we prove in general that if:
$$ f^{(n)}(x) = (-1)^n \frac{(n+1)!}{(1+x)^{n+2}}$$
Then we verify by differentiating this that:
$$ f^{(n+1)}(x) = (-1)^{n+1} \frac{(n+2)!}{(1+x)^{n+3}}$$
So now we derive the series from taylors formula, 
$$ f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \frac{1}{3!}f'''(a)(x-a)^3 + ... $$
Letting $a = 0$
$$ f(x) = f(a) + f'(a)x + \frac{1}{2!}f''(a)x^2 + \frac{1}{3!}f'''(a)x^3 + ... $$
Now substituting:
$$ \frac{1}{(1+x)^2} = 1-\frac{2}{(1+0)^3}x -  \frac{1}{2!}\frac{3!}{(1+0)^4}x^2 + \frac{1}{3!} \frac{4!}{(1+0)^5}x^3 + ...$$
And simplifying $(1+0) = 1, \frac{(n+1)!}{n!} = (n+1)$ we have
$$ \frac{1}{(1+x)^2} = 1-2x +  3x^2 - 4x^3 + ...$$
As desired
