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.
If we ignore statement about monotonic decreasing function, isn’t it a definition of integral convergence?
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.
It seems easier to prove by directly using the definition of the limit.
…but the problem with my suggestion is that one limit uses integers and the other one uses reals and we should first bridge the gap (potentially the way Evangelos did)…so I take back the above comment.
Wouldn’t f(x) = 1/x work as a counterexample here? I don’t think we can say the intgral is convergent for all x since we don’t have a bound on the I_n (unless I misread the question and the limit can’t be infinite, in which case I think this would work and is a smart way to do it!)
Yes, it works