How can I find this sum? 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?


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.

