# Summation evaluation

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

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

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 ?