Changing summation in a power series I'm doing a question in my power series unit that involves adding summations together, I just started this unit so I'm not totally clear on how changing summation works, from what I understand you want to make the x term in the summation into $x^n$. From what I've seen in lessons it seems the summation is changed depending on what that change in n is.
The problem I am stuck on is this:
$2\sum_{n=0}^∞ a_nx^{n+1} +\sum_{n=1}^∞ nb_nx^{n-1}$
Using my understanding of the unit I did this to set $x^n$ :
$2\sum_{n=1}^∞ a_{n-1}x^{n} +\sum_{n=0}^∞ (n+1)b_{n+1}x^{n}$
Up until now when I change the Xs to $x^n$ I have been able to put the two summations into one because they end up being the same n, but here the summations are different so I can't put them together and take out $x^n$ to move onto the next step, which tells me that my understanding of the summation change so far has only worked coincidentally, I'm pretty sure I'm not understanding how changing the Ns and summations work, can someone explain how the changing Ns and the summation are related?  
Because if the summation changes every time I change the N then I would never be able to add these two summations together for this problem, which is obviously not true. I'm not interested in the answer, I just want to know how this specific step works where you change the summations to the same value and still have the $x^n$ in both summations.
Just to clarify my understanding of the summation change, this is how I think it works at this moment: 
if I have a problem like 
$\sum_{n=1}^∞ a_nx^{n-1}$
I see that the exponent on x is n-1, so I set n=n+1, which changes all ns to n+1 which changes the n in the summation to n=0.
which would get me the result
$\sum_{n=0}^∞ a_{n+1}x^{n}$
but it turns out its more complicated then this since it didn't work with the more complicated questions like this one I ran into. 
 A: Reindexing a sum in this way is fine.
Pulling the first few terms in/out of the infinite sum is also fine.  That is $\sum_{n=0}^\infty a_n = a_0 + \sum_{n=1}^\infty a_n$.
The trickiest part of what you are doing is that combining terms of an infinite sum piecewise is fine if both sums are well-defined.  That is if $\sum_{n=k}^\infty a_n$ and $\sum_{n=k}^\infty b_n$ both are finite numbers, then $\sum_{n=k}^\infty a_n + \sum_{n=k}^\infty b_n = \sum_{n=k}^\infty (a_n+b_n)$.  This follows from the fact that if $N$ is large then $$\sum_{n=k}^\infty (a_n+b_n) \approx \sum_{n=k}^N (a_n+b_n) = \sum_{n=k}^N a_n + \sum_{n=k}^N b_n \approx \sum_{n=k}^\infty a_n + \sum_{n=k}^\infty b_n$$
(There are three approximations.  For any $\epsilon > 0$ you can make $N$ large enough to make all three within $\frac{\epsilon}{3}$.  You can turn that into an official proof very easily.)
A: You could calculate as follows

\begin{align*}
2\sum_{n=0}^\infty& a_nx^{n+1} +\sum_{n=1}^ \infty nb_nx^{n-1}\\
&=2\sum_{n=1}^\infty a_{n-1}x^{n} +\sum_{n=0}^ \infty (n+1)b_{n+1}x^{n}\tag{1}\\
&=2\sum_{n=1}^\infty a_{n-1}x^{n} +\left(b_1+\sum_{n=1}^ \infty (n+1)b_{n+1}x^{n}\right)\tag{2}\\
&=b_1+\sum_{n=1}^{\infty}\left(2a_{n-1}+(n+1)b_{n+1}\right)x^n\tag{3}
\end{align*}

Comment:


*

*In (1) we shift the index of both series by one in order to obtain summands with $x^n$.

*In (2) we separate the first summand of the second series. So we get the same index starting value $n=1$ in both series.

*In (3) we collect the terms into one series.

If we denote the series (3) with
  \begin{align*}
C(x)=\sum_{n=0}^\infty c_nx^n
\end{align*}
  we observe
  \begin{align*}
c_0&=b_1\\
c_n&=2a_{n-1}+(n+1)b_{n+1}\qquad\qquad n\geq 1
\end{align*}

