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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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This is not a typo, you meant $\int_0^1|f'(x)|dx$ ? – Norbert Apr 7 '12 at 23:15
It's definitely a typo, as $f(t)=|\sin 2\pi t|$ would give us $0$ for the series and $\infty$ for the integral. – Martin Argerami Apr 7 '12 at 23:21
@Norbert : You're right! I fixed the statement now, it should be: "$\int_{1}^{\infty }\left | f^{'}(x) \right |dx$ converges" instead of "$\int_{0}^{1 }\left | f^{'}(x) \right |dx$ converges" – M.Krov Apr 7 '12 at 23:34
Can anyone provide a solution to this problem? – M.Krov Apr 8 '12 at 2:10
By converges, do you imply bounded? – Suresh Apr 8 '12 at 4:31
up vote 5 down vote accepted

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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@ user15464: I don't understand how you derived the equalities: $\int_n ^{n+1}\int_n ^x |f'(t)| dt = \int_{n} ^{n+1} \int_t ^{n+1} |f(t)| dt = \int_{n} ^{n+1} (n+1 - t)|f(t)| dt \leq \int_{n} ^{n+1} |f(t)| dt$. Can you, please, do it step by step so that I can understand what you're doing? Thanks – M.Krov Apr 8 '12 at 4:20
@M.Krov: It should be $\int_n^{n+1}\int_n^x|f'(t)|dtdx=\int_n^{n+1}\int_t^{n+1}|f'(t)|dxdt=$ $\int_n^{n+1}(n+1-t)|f'(t)|dt\le\int_n^{n+1}|f'(t)|dt$; it’s just interchanging the order of integration, with a lot of typos. I’m going to take the liberty of editing the answer to fix it. – Brian M. Scott Apr 8 '12 at 5:24
@Brian M. Scott: Can you please show me how you proved the inequality: $$\int_{n}^{n+1}(n+1-t)\left | f^{'}(t) \right |dt\leqslant \int_{n}^{n+1}\left | f^{'}(t)\right |dt?$$? Obviously, $(n+1-t)$ is positive, but it has to be less than $1$ for the inequality to be true. – M.Krov Apr 8 '12 at 6:23
@M.Krov: As $t$ ranges from $n$ to $n+1$, $n+1-t$ ranges from $1$ to $0$, so $(n+1-t)|f'(t)|\le|f'(t)|$ over the whose interval $[n,n+1]$. – Brian M. Scott Apr 8 '12 at 6:25
@Brian M. Scott: You're right, I didn't pay attention that $t$ is bigger than $n$. Thanks – M.Krov Apr 8 '12 at 6:28

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