Given $f(k)=(0.5)^k$ and $g(k)=(0.25)^k$, I wanted to find $T(8)$ using convolution where $T(k)=f(k)*g(k)$. I did following calculation but not sure everything is in place as I am getting final answer as infinity.
So, $T(k)=f(k)*g(k)=\int_{-\infty}^{\infty}f(x)g(k-x)dx$
Therefore, $T(8)=\int_{-\infty}^{\infty}f(x)g(8-x)dx=\int_{-\infty}^{\infty}(0.5)^x(0.25)^{8-x}dx=\frac{1}{2^{16}}\int_{-\infty}^{\infty}2^xdx=\frac{1}{2^{16}\log2}[2^x]_{-\infty}^{\infty}$, which is divergent.