Power series $f(x)=\frac1{(1+x)^2}$ I understand that the answer is $$-\sum _{n=1}^{\infty}(-1)^nn\cdot x^{n-1}$$
As shown in this answer here
But why does the n have to start at n=1? Why can it not start at any number? Also how would I write it in the form where n=0? The answer is $$\sum_{k=0}^{\infty}\left(-1\right)^k\left(k+1\right)x^k$$ but what happened to the negative sign before the $\sum$?
 A: 
But why does the n have to start at n=1? Why can it not start at any number?

The answer to the problem ultimately involves differentiation of the geometric series. We know that
$$1+x+x^2+x^3+\cdots = \sum_{n=0}^\infty x^n = \frac{1}{1-x}$$
provided $|x|<1$. If you were to take the derivative of this summation and $1/(1-x)$, you are allowed to do so term-by-term in the summation since it is absolutely convergent. However, that means
$$1+x+x^2+x^3+\cdots \overset{d/dx}{\implies} 0 + 1 + 2x + 3x^2 + \cdots$$
The $n=0$ term, which was $1$, is $0$ in the derivative. Thus we can skip over the $n=0$ index altogether and just start at $n=1$. Thus we claim
$$\frac{d}{dx} \left( \sum_{n=0}^\infty x^n \right) = \sum_{n=1}^\infty nx^{n-1}$$
Indeed, this expression makes it equally clear: if you plug in $n=0$ to the right summation, you get a redundant $+0$ summand. 
Thus, in summary:


*

*We start the geometric series at $n=0$ to make use of the known result.

*The $n=0$ term becomes a $0$ in the derivative. We thus, effectively, chop it off and start at $n=1$ instead. It's a simplification in a sense, in other words.




Also how would I write it in the form where n=0? The answer is $$\sum_{k=0}^{\infty}\left(-1\right)^k\left(k+1\right)x^k$$ but what happened to the negative sign before the $\sum$?

(Minor note, you had $n$ in your original summation in the quoted bit. I fixed it to have $k$ here.)
Distribute in the negative sign term by term: since the summation converges absolutely, we can claim $c \sum a_k = \sum ca_k$ after all:
$$-\sum_{k=0}^{\infty}\left(-1\right)^k\left(k+1\right)x^k = \sum_{k=0}^{\infty}\left(-1\right)^{k+1}\left(k+1\right)x^k$$
You can then re-index the summation by defining $n=k+1$. Then...


*

*Lower bound: $k=0$ gives $n=1$

*Upper bound: Obviously $k \to \infty$ gives $n \to \infty$

*Summands: Since $n=k+1$, we can replace those quite easily, and $x^k$ becomes $x^{n-1}$.


Thus,
$$\sum_{k=0}^{\infty}\left(-1\right)^{k+1}\left(k+1\right)x^k = \sum_{n=1}^{\infty}\left(-1\right)^{n}nx^{n-1}$$
Thus, your first expression from your post, and the second answer given to you, are equivalent.
A: It is a change of index $n\mapsto k+1$: $$\sum_{n=1}^\infty a_n = \sum_{k+1=1}^\infty a_{k+1} = \sum_{k=0}^\infty a_{k+1}$$
A: The negative sign is moved into the series by changing the exponents of $-1$. Try to write out the first few terms to see what happens.
A: You mix up $k$ and $n$ in your second formula. A correct expression there would be
$\sum_{n=0}^\infty (-1)^n (n+1) x^n $
or
$\sum_{k=0}^\infty (-1)^k (k+1) x^k $.
Because if we take the second one, and let $n=k+1$, we get
$\sum_{n=1}^\infty (-1)^{n-1} \cdot n \cdot x^{n-1} $;
now use
$(-1)^{n-1}= -(-1)^n$.
A: $$\frac1{1+x}=\sum_{n=0}^\infty (-x)^n\Longrightarrow -\frac1{(1+x)^2}=-\sum_{n=0}^\infty n(-x)^{n-1} $$
Clearly the first term is zero due to the coefficient of $n$ so $n$ can start from $1$
$$-\frac1{(1+x)^2}=-\sum_{n=1}^\infty n(-x)^{n-1}= -\sum_{n=0}^\infty (n+1)(-x)^{n}$$
Last step follows from using the rule  $$\sum_{n=a}^\infty f(n)=\sum_{n=a-1}^\infty f(n+1)$$
