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

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.

One comment

  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.

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