Question 1184309


Let k = 1, 2, 3, 4, 5, ...

Then for an infinite geometric series with {{{g[1] = k}}} and {{{r = k/(k+1) <1}}}, then for {{{abs(r) < 1}}},

{{{S = g[1]/(1-r) = k/(1-k/(k+1)) = k/(1/(k+1)) = k(k+1)}}}


===> {{{lim(k->infinity, (-1)^k/S) =lim(k->infinity, (-1)^k/(k(k+1))) = highlight(0)}}}