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.

  • $\begingroup$ Changing the $n$ index to $0/1/-1$ doesn't matter,you can do it in any question regardless of the context. In your case, make it like:$$2\sum_{n=1}^\infty a_{n-1}x^n + \sum_{n=1}^\infty nb_nx_{n-1}$$, this does not change the problem, makes the indices same so that later you can combine the summations if the power series are absolutely convergent. $\endgroup$ Commented Apr 5, 2016 at 0:17
  • $\begingroup$ So I can change my summation to whatever I want and it won't affect my equation in the summation at all? $\endgroup$
    – kek
    Commented Apr 5, 2016 at 0:19
  • 1
    $\begingroup$ No. The index is of no consequence whatsoever. It doesn't make a difference. What should be your question is : when can I combine the summations? That, I say, is under absolute convergence. As an exercise, put $n=-3$ in both and form different summations: The terms inside are not changing, so the summation has no reason to change. $\endgroup$ Commented Apr 5, 2016 at 0:20
  • $\begingroup$ You need to combine terms with same power of x - note that the index need to fit smaller power. $\endgroup$
    – Moti
    Commented Apr 5, 2016 at 1:27
  • $\begingroup$ @астонвіллаолофмэллбэрг Absolute convergence is not needed in this case. Convergence suffices. You would need absolute convergence for infinite rearrangement of terms. But not piecewise combination of them. $\endgroup$
    – btilly
    Commented Apr 5, 2016 at 1:30

2 Answers 2


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*}


  • 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*}


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.)

  • $\begingroup$ What you are saying is correct.$+1$, and apologies for my mistake earlier. $\endgroup$ Commented Apr 5, 2016 at 1:33

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