Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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?

share|cite|improve this question
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 $x\in [n, n+1]$.

share|cite|improve this answer
@ 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.