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

by RoRi

The following function f is defined for all x \geq 1, and n is a positive integer. Prove or provide a counterexample to the following statement.

If f is monotonically decreasing and if

    \[ \lim_{n \to \infty} I_n \]

exists, then the improper integral

    \[ \int_1^{\infty} f(x) \, dx \]

converges.

Incomplete.

4 comments

  1. Evangelos says:

    I think this proof holds… Would be happy to see any counterexamples.

    Proof. If f is monotonic decreasing and the limit of I_{n} 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 n \leq x \leq n + 1,

        \begin{align*} \int_{1}^{x} f(t) \, dt \leq I_{n+1} \\ \leq I_{n} + f(n) \end{align*}

    But since f(n) goes to zero as n goes to infinity, \int_{1}^{x} f(t) \,dt is thus bounded by I_{n} for all n \leq x \leq n+1 as n goes to infinity. Thus, \int_{1}^{x} f(t) \,dt is convergent for all x as x goes to infinity.

      • 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!)

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