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.**

Skip to content
#
Stumbling Robot

A Fraction of a Dot
#
Prove or disprove: If *f* is monotonic decreasing and *lim I*_{n} exists then *∫ f(x)* converges

###
4 comments

### Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):

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.

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!)