If $a_n$ → a, $b_n$ → b, then $c_n$ = ($a_1$$b_n$+$a_2$$b_{n−1}$+···+$a_n$$b_1$)\n → ab.
Hint: Write $a_n$ = a + $\epsilon$$_n$, where $\epsilon$ → 0. Then:
$c_n$ = ($a_1$$b_n$+$a_2$$b_{n−1}$+···+$a_n$$b_1$)\n = a [($b_n$ + $b_{n−1}$ + ··· + $b_1$)/n]+[($\epsilon_1$$b_n$ + $\epsilon_2$$b_{n-1}$ + ··· + $\epsilon_n$$b_1$)/n]
The Questions:
The first part tends to ab (why?), and it remains to show that the second part tends to zero