Where is the mistake in the convolution? Let :
$$
x[n]=\begin{cases}
1&amp;\text{if $0\leq n\leq 4$}\\ \\
0&amp;\text{if otherwise}
\end{cases}
\qquad
\text{and}
\qquad
h[n]=\begin{cases}
\alpha^{n}&amp;\text{if $0\leq n\leq 6$}\\ \\
0&amp;\text{if otherwise}
\end{cases}
$$
We must compute the convolution :
$$
(x*h)[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]
$$
Observe that $x[n]=u[n]$ for $n\in[0,4]$ and $0$ otherwise. Furthermore, $h[n]=\alpha^{n}u[n]$ for $n\in[0,6]$. Thus :
\begin{array}{c | c c c c c c c }
\mathbf{x} &amp; 1 &amp; \alpha &amp; \alpha^{2} &amp; \alpha^{3} &amp;\alpha^{4} &amp;\alpha^{5} &amp;\alpha^{6}\\   
\hline 
1 &amp; 1 &amp; \alpha &amp; \alpha^{2} &amp; \alpha^{3} &amp;\alpha^{4} &amp;\alpha^{5} &amp;\alpha^{6}\\  
1 &amp; 1 &amp; \alpha &amp; \alpha^{2} &amp; \alpha^{3} &amp;\alpha^{4} &amp;\alpha^{5} &amp;\alpha^{6}\\    
1 &amp; 1 &amp; \alpha &amp; \alpha^{2} &amp; \alpha^{3} &amp;\alpha^{4} &amp;\alpha^{5} &amp;\alpha^{6}\\    
1 &amp; 1 &amp; \alpha &amp; \alpha^{2} &amp; \alpha^{3} &amp;\alpha^{4} &amp;\alpha^{5} &amp;\alpha^{6}\\  
1 &amp; 1 &amp; \alpha &amp; \alpha^{2} &amp; \alpha^{3} &amp;\alpha^{4} &amp;\alpha^{5} &amp;\alpha^{6}\\     
0 &amp; 0 &amp; 0 &amp; 0&amp; 0&amp; 0 &amp; 0 &amp; 0 \\  
0 &amp; 0 &amp; 0 &amp; 0&amp; 0&amp; 0 &amp; 0 &amp; 0
\end{array}
\begin{align*}
\displaystyle (x*h)[n]&amp;=\left[\sum_{k=0}^{0}\alpha^{k}\;,\;\sum_{k=0}^{1}\alpha^{k}\;,\;\cdots\;,\;\sum_{k=0}^{4}\alpha^{k}\;,\;\sum_{k=1}^{5}\alpha^{k}\;,\;\sum_{k=2}^{6}\alpha^{k}\;,\;\cdots\;,\;\sum_{k=5}^{6}\alpha^{k}\;,\;\sum_{k=6}^{6}\alpha^{k}\right]\\ \\
&amp;=\begin{cases}
\displaystyle\sum_{k=0}^{n}\alpha^{k}&amp;\text{if $0\leq n\leq 4$}\\
\displaystyle\sum_{k=1}^{n}\alpha^{k}&amp;\text{if $n=5$}\\
\displaystyle\sum_{k=n+2}^{6}\alpha^{k}&amp;\text{if $0\leq n\leq4$}
\end{cases}
\end{align*}
I wish to know where my mistake is and how to fix it. Thank you :)
$3^{rd}$ term in your final answer should have $6 \le n \le 10$, instead of $0 \le n \le 4$. And, the limit of summation will go from $n-4$ to $6$.
$$\sum^{6}_{k = n-4} \alpha^k \ ,  6 \le n \le 10$$
$0 \le n \le 4$ case has already been covered by the $1^{st}$ term of final answer.