The following function is defined for all
, and
is a positive integer. Prove or provide a counterexample to the following statement.
If is monotonically decreasing and if
exists, then the improper integral
converges.
Incomplete.
I think this proof holds… Would be happy to see any counterexamples.
Proof. If f is monotonic decreasing and the limit of
exists as n goes to infinity, then f must go to zero as n goes to infinity. And since f is monotonic decreasing for all x, for any
,
But since f(n) goes to zero as n goes to infinity,
is thus bounded by
for all
as n goes to infinity. Thus,
is convergent for all x as x goes to infinity.