How do you find the sum of a series going from negative infinity to -1?

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Going from $\sum_{n=-\infty}^{-1} 2^n z^{-n}$ to $\sum_{k=1}^\infty 2^{-k} z^{k}$ is just a substitution.

Start with $\sum_{n=-\infty}^{-1} 2^n z^{-n}$.

Let $k = -n$. Then when $n = -\infty$, we have $k = -n = -(-\infty) = +\infty$. Also, when $n = -1$, we have $k=-n = -(-1) = 1$. So the limits of the new sum in terms of $k$ are $1$ and $\infty$.

But why did the limits change position? That is, why is the "infinity" limit on the bottom for the $n$ sum but the corresponding "infinity" limit is on the top for the $k$ sum? Addition is commutative, so it doesn't matter what order we add things in, and it's custom to put the smaller sum limit on the bottom of the $\sum$. This is why we say $\sum_{k=1}^{\infty}$ instead of $\sum_{k=\infty}^1$. Actually, the notation $\sum_{k=\infty}^1$ doesn't have a universally agreed upon interpretation if I recall correctly. But in this context we know it's safe to say $\sum_{k=1}^{\infty}$ because it came from us rewriting the existing sum $\sum_{n=-\infty}^{-1}$.

So what about the $2^nz^{-n}$? Recall that we said $k = -n$. So then $n = -k$. This means $2^n = 2^{-k}$ and $z^{-n} = z^{-(-k)} = z^k$. Therefore $2^nz^{-n} = 2^{-k}z^k$ and all together we get $\sum_{k=1}^\infty 2^{-k} z^{k}$.

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atosh502
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Updated on December 20, 2022

Comments

  • atosh502
    atosh502 11 months

    The series is : $\sum_{n=-\infty}^{-1} 2^n z^{-n}$

    I found that you can invert the series to this: $\sum_{k=1}^\infty 2^{-k} z^{k}$

    I want to know how it is possible and some explanation.

    • Elad
      Elad over 6 years
      This sum is a limit. Now use the delta epsilon formulation to observe for yourself that both limits are the same. You won't have to handle infinity this way.
  • badjohn
    badjohn over 6 years
    Despite addition bring commutative, the order can matter in an infinite sum. It might be conditionally but not absolutely convergent.