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# Prove or disprove: If f is monotonic decreasing and lim In exists then ∫ f(x) converges

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.

1. Anonymous says:

If we ignore statement about monotonic decreasing function, isn’t it a definition of integral convergence?

2. Evangelos says:

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.

• S says:

It seems easier to prove by directly using the definition of the limit.

• S says:

…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.

• William C says:

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!)