I did see this discussed in another question; however, I did exactly what the answer suggested and still do not seem to get the solution.
The question is to use differentiation to find the representative power series of $1/(x+1)^2$
I could easily see that the derivative of $-1/(x+1)$ is $1/(x+1)^2$.
So, my goal was to find the power series for $-1/(x+1)$ and then take the derivative of that.
I found the power series for $-1/(x+1)$ to be: $\sum_{n=0}^\infty (-1)^\left(n+1\right) x^n$.
Then I simply took the derivative of that power series to get:
$\sum_{n=0}^\infty (n) (-1)^\left(n+1\right) x^\left(n-1\right)$
Can't see where I went wrong.
Book answer: $\sum_{n=0}^\infty (-1)^n (n+1)x^n$
Makes no sense to me. Maybe I didn't find the original power series correctly?
Best Answer
It's the same. In your answer, the $n=0$ term is just $0$ because of the multiplication by $n$: \begin{align} \sum_{n=0}^{\infty}n(-1)^{n+1} x^n = \sum_{n=1}^{\infty}n(-1)^{n+1}x^n. \end{align} Now, change the index of summation, by letting $n=m+1$. Then, \begin{align} \sum_{n=1}^{\infty}n(-1)^{n+1}x^n=\sum_{m+1=1}^{\infty}(m+1)(-1)^{m+1+1}x^{m+1}= \sum_{m=0}^{\infty}(-1)^m(m+1)x^{m+1} \end{align} Now, you can relabel the indices however you like: \begin{align} \sum_{m=0}^{\infty}(-1)^m(m+1)x^{m+1}=\sum_{n=0}^{\infty}(-1)^n(n+1)x^{n+1}= \sum_{@=0}^{\infty}(-1)^@(@+1)x^{@+1}= \sum_{\star=0}^{\infty}(-1)^{\star}(\star+1)x^{\star+1}. \end{align}