If n takes values less than $-1$ how do I evaluate the following summation:

$$y[n]= \sum_{k=n+1}^{\infty} a^{n-k}$$

Do I need to switch signs of n because n takes only negative values? I'm sorry if this question seems stupid....I'm not that well versed with summation

Edit: a is a number greater than 1

The way I tried to solve it is:

$$y[n]= a^n((\sum_{k=n+1}^{-1} a^{-k})+(\sum_{k=0}^{\infty} a^{-k}))$$

where the second summation evaluates to $$\frac{1}{1-a^{-1}}$$

but somehow the first summation seems to give me a problem

  • 1
    $\begingroup$ It needs more information. What is $a$ here ? What values can $n$ take ? $\endgroup$
    – JRC
    Nov 16, 2020 at 9:32
  • $\begingroup$ HINT: $a^{n-k}=\dfrac{1}{a^{k-n}}$ $\endgroup$ Nov 16, 2020 at 9:32
  • $\begingroup$ @Kolmogorov: $n<-1$, no ? $\endgroup$
    – user65203
    Nov 16, 2020 at 10:00
  • $\begingroup$ Yes @Yves Daoust n<-1 $\endgroup$
    – Orpheus
    Nov 16, 2020 at 10:02

3 Answers 3


As $a^{n-k}=\dfrac{1}{a^{k-n}}$, your sum is $$y[n]= \sum_{k=n+1}^{\infty} a^{n-k}=\sum_{k-n=1}^{\infty}\left(\frac{1}{a}\right)^{k-n}=\sum_{m=1}^{\infty}\left(\frac{1}{a}\right)^{m}$$ (where in the last equality i've change $k-n$ by $m$). So you have a geometric series (independent of $n$) and you probably can solve it by your own.


You evaluate it as it is: as $\lim_{h\to\infty}S_h$, where $S_\bullet$ is the sequence $S_\bullet:\{x\in\Bbb Z\,:\, x\ge n+1\}\to\Bbb R$ such that $S_{n+1}=a^{n-(n+1)}$ and, $\forall m\in\{x\in \Bbb Z\,:\, x\ge n+1\}$, $S(m+1)=a^{n-m-1}+S_m$.


Whatever the value of $n$ your sum is

$$a^{-1}+a^{-2}+a^{-3}+\cdots$$ so why worry ?


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