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Prove or disprove: ∫ f(x) converges implies lim f(x) = 0

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.

The convergence of the improper integral

    \[ \int_1^{\infty} f(x) \, dx \]

implies

    \[ \lim_{x \to \infty} f(x) = 0. \]


Counterexample. The idea of the construction is a function which has rapidly diminishing area, but has a height that is not going to 0. (So, for an idea consider triangles on the real line all with height 1, but for which the base is becoming small rapidly.) To make this concrete, define

    \[ f(x) = 1 - 2n^2 \left| x - \left( n + \frac{1}{2n^2} \right) \qquad \text{for } x \in [n,n+1), \]

for each positive integer n. Then for the improper integral we have

    \[ \int_1^{\infty} f(x) \, dx \leq \sum_{n=1}^{\infty} \frac{1}{2n^2} \]

which we know converges. On the other hand

    \[ f \left( n + \frac{1}{2n^2} \right) = 1 \]

for all positive integers n. Hence,

    \[ \lim_{x \to \infty} f(x) \neq 0 \]

(since it does not exist). Hence, the statement is false.

(Note: For more on this see this question on Math.SE.)