How can I find this sum?

Question

I'm doing some examples related to convolution (digital signal processing). I post my problem here because it is actually mathematics problem.

I have to calculate this sum:

$$\sum_{k\ = \ n-5}^{n+5} e^{-|k|}$$

Any suggestion?

Answer 1

We show the following is valid: \begin{align*} \sum_{k=n-5}^{n+5}e^{-|k|}= \begin{cases} {\displaystyle \frac{e^{-5}-e^{6}}{1-e}e^{-n}}&\qquad\qquad |n|>5\\ \\ {\displaystyle \frac{e^{-n-5}-e^{n-5}-1-e}{1-e}}&\qquad\qquad |n|\leq 5 \end{cases} \end{align*}

Since \begin{align*} e^{-|k|}= \begin{cases} e^{-k}&\qquad k\geq 0\\ e^k&\qquad k<0 \end{cases} \end{align*}

we distinguish three different cases:

\begin{align*} \sum_{k=n-5}^{n+5}e^{-|k|}= \begin{cases} {\displaystyle\sum_{k=n-5}^{n+5}e^{k}}&\qquad n<- 5\qquad \qquad\text{(i)}\\ {\displaystyle\sum_{k=n-5}^{-1}e^{k}+\sum_{k=0}^{n+5}e^{-k}}&\qquad -5\leq n\leq 5\qquad \text{(ii)}\\ {\displaystyle\sum_{k=n-5}^{n+5}e^{-k}}&\qquad n> 5\ \ \qquad \qquad\text{(iii)}\\ \end{cases} \end{align*}

Note that cases (i) and (iii) coincide, since substituting $k$ with $-k$ and changing the order of summation in (iii) gives \begin{align*} \sum_{k=n-5}^{n+5}e^{-k}=\sum_{-k=n-5}^{n+5}e^{k}=\sum_{k=-n-5}^{-n+5}e^{k} \end{align*} which is the same as (i) when $n$ is replaced is with $-n$.

We recall the formula of finite geometric series and do some index transformation \begin{align*} \sum_{k=0}^mq^k&=\frac{1-q^{m+1}}{1-q}\\ \sum_{k=a}^bq^k&=\sum_{k=0}^{b-a}q^{k+a} =q^a \sum_{k=0}^{b-a}q^{k}\\ &=q^a\frac{1-q^{b-a+1}}{1-q} \end{align*}

Now we are well prepared to calculate the cases (i) - (iii)

Case (i): $n<-5$ \begin{align*} \sum_{k=n-5}^{n+5}e^{-|k|}&=\sum_{k=n-5}^{n+5}e^{k}\\ &=\frac{1-e^{[n+5-(n-5)+1]}}{1-e^{-1}}e^{n-5}\\ &=\frac{1-e^{11}}{1-e}e^{n-5}\\ &=\frac{e^{-5}-e^{6}}{1-e}e^{n}\\ \end{align*} Case (ii): $-5\leq n\leq 5$ \begin{align*} \sum_{k=n-5}^{n+5}e^{-|k|}&=\sum_{k=n-5}^{-1}e^{k}+\sum_{k=0}^{n+5}e^{-k}\\ &=\sum_{k=0}^{-n+4}e^{k+n-5}+\frac{1-e^{-[(n+5)+1]}}{1-e^{-1}}\\ &=e^{n-5}\frac{1-e^{[(-n+4)+1]}}{1-e}+\frac{1-e^{-n-6}}{1-e^{-1}}\\ &=e^{n-5}\frac{1-e^{-n+5}}{1-e}+\frac{1-e^{-n-6}}{1-e^{-1}}\\ &=\frac{e^{-n-5}-e^{n-5}-1-e}{1-e} \end{align*} Case (iii): $n>5$ \begin{align*} \sum_{k=n-5}^{n+5}e^{-|k|}&=\sum_{k=n-5}^{n+5}e^{-k}\\ &=\frac{1-e^{-[n+5-(n-5)+1]}}{1-e^{-1}}e^{-(n-5)}\\ &=\frac{1-e^{-11}}{1-e^{-1}}e^{-n+5}\\ &=\frac{e^5-e^{-6}}{1-e^{-1}}e^{-n}\\ &=\frac{e^{-5}-e^{6}}{1-e}e^{-n}\\ \end{align*}

and the claim follows.