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Problem:

Let $f\in C^{1}([0,\infty ))$ such that: \int_{1}^{\infty }\left | f^{'}(x) \right |dx converges. The question is to prove the following:

$\left ( \sum_{n=1}^{\infty }f(n) \right )$ converges $\Leftrightarrow \left ( \int_{1}^{\infty }f(x)dx \right )$ converges

I don't know how to prove it. For the direction: $\Leftarrow $ I was trying to use the definition of Rieamann integrals as an infinite sum where the mesh goes to zero, and somehow try to prove that $\left ( \sum_{n=1}^{\infty }f(n) \right )$ converges.

Any solution or ideas for this problem?

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    By converges, do you imply bounded?2012-04-08

1 Answers 1

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Suppose first that the sum converges. By the fundamental theorem of calculus, for each $x\in [n,n+1]$,

f(x) = f(n) + \int_{n} ^x f'(t) dt

and therefore

|f(x)| \leq |f(n)| + \int_n ^x |f'(t)| dt\;.

Integrating and summing over all $n$, we get

\begin{align*} \int_1 ^\infty |f(x)| dx &= \sum_{n=1} ^\infty \int_n ^{n+1} |f(x)| dx\\ &\leq \sum_{n=1} ^\infty |f(n)| + \sum_{n=1} ^\infty \int_n ^{n+1}\int_n ^x |f'(t)| dt dx \;.\tag{1}\end{align*}

We have

\begin{align*} \int_n ^{n+1}\int_n ^x |f'(t)| dtdx &= \int_{n} ^{n+1} \int_t ^{n+1} |f'(t)| dxdt \\ &= \int_{n} ^{n+1} (n+1 - t)|f'(t)| dt \\ &\leq \int_{n} ^{n+1} |f'(t)| dt\;.\end{align*}

So, the final sum in $(1)$ is at most \sum_{n=1} ^\infty \int_n ^{n+1}|f'(t)| dt = \int_1 ^\infty |f'(t)| dt\;. Thus

\int_1 ^\infty |f(x)| dx \leq \sum_{n=1} ^\infty|f(n)| + \int_1 ^\infty |f'(t)| dt < \infty\;.

For the converse, we use a similar argument, beginning from the equation

f(n) = f(x) - \int_{n} ^x f'(t) dt

for each $x\in [n, n+1]$.

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    @Brian M. Scott: You're right, I didn't pay attention that $t$ is bigger than $n$. Thanks2012-04-08