Consider the following sequences:
$x_{1}(n) = A$ (a constant), $x_{2}(n) = u(n)$, $x_{3}(n) =\delta(n)-\delta(n-1)$.
($\circledast$ stands for linear convolution)
If I perform the operation $x_{2}\circledast (x_{3}\circledast x_{1})$, the value I am getting is $0$, where as if I perform $(x_{2}\circledast x_{3})\circledast x_{1}$ the value I am getting is A.
Since Convolution is associative, why are the answers different?
My approach: In general, assuming $x_{1}(n)$, $x_{2}(n)$, $x_{3}(n)$ exists from $-\infty$ to $\infty$,
$$x_{2}\circledast (x_{3}\circledast x_{1})=\sum_{k=-\infty}^{\infty}x_2(k).x_{3}\circledast x_{1}(n-k)=\sum_{k=-\infty}^{\infty}x_2(k)\sum_{l=-\infty}^{\infty}x_3(l)x_1(n-k-l)$$
Let $m=n-k-l$,
$$=\sum_{k=-\infty}^{\infty}\sum_{l=-\infty}^{\infty}x_2(k)x_3(l)x_1(n-k-l)=\sum_{k=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}x_2(k)x_3(n-m-k)x_1(m)$$
$$=\sum_{m=-\infty}^{\infty}x_{2}\circledast x_{3}(n-m)x_1(m)=(x_{2}\circledast x_{3})\circledast x_{1}$$
Hence I feel Convolution is associative even if the functions are of infinite lengths. Where did I go wrong?
You have skipped one step, the one in red: $$ \sum_{k=-\infty}^{\infty} \sum_{m=-\infty}^{\infty} x_2(k) \, x_3(n-m-k) \, x_1(m) \\ = \color{red}{ \sum_{m=-\infty}^{\infty} \sum_{k=-\infty}^{\infty} x_2(k) \, x_3(n-m-k) \, x_1(m) } \\ = \sum_{m=-\infty}^{\infty} x_{2} \circledast x_{3}(n-m) \, x_1(m) $$
Under what conditions are you allowed to swap the order of summations? Are they fulfilled here?