MATH 101 V2A
February 25th – Practice problems
Solutions
1. Determine whether the following statements are true or false. If true, provide justification. If false,
provide a counterexample.
(a) If {an } diverges to infinity and {bn } converges to 0, then {an bn } diverges to infinity.
Solution: This is false! Intuitively, it should be false for the same reason the analogous statement
about functions (if limx→∞ f (x) = ±∞ and limx→∞ g(x) =0, then limx→∞ f (x)g(x) = ±∞) is
1
. Then {an } diverges to infinity
false. For a counterexample, let {an } ={n}
and
let
{b
}
=
2
n
n
and {bn } converges to 0, but {an bn } = n1 also converges to 0.
(b) If {an } diverges to infinity and {bn } converges to 1, then {an bn } diverges to infinity.
Solution: This is true. Intuitively, if {bn } converges to 1, then for large n, an bn ≈ an , and an will
be getting arbitrarily large. Now let’s justify this more carefully. Assume {an } diverges to +∞
(the case when {an } diverges to −∞ is similar). If {bn } converges to 1, then there is an integer
N > 0 such that, for all n ≥ N , (choosing ε = 21 )
1−
Therefore, for all n ≥ N , an bn >
infinity as well.
an
2 .
1
1
< bn < 1 + .
2
2
So, since limn→∞
an
2
= ∞, we see that {an bn } diverges to
(c) If {an } and {bn } both diverge, then {an bn } also diverges.
Solution: This is false! Notice that the statement does NOT say that either sequence diverges to
infinity. For a counterexample, let {an } = {(−1)n } = {bn }. Then {an } and {bn } both diverge
(but not to infinity), but {an bn } = {1}, which clearly converges to 1.