# $\int_{1}^{\infty}f$ converges if and only if $\sum_{n=1}^{\infty}\int_{n}^{n+1}f$ converges

In order to prove the integral test for convegence, I want to show that if $f:[1,\infty)\to\mathbb{R}$ is nonnegative and monotone decreasing then:

$\int_{1}^{\infty}f$ converges if and only if $\sum_{n=1}^{\infty}\int_{n}^{n+1}f$ converges

Clearly, since $f$ is nonnegative if $\int_{1}^{\infty}f$ converges then $\sum_{i=1}^n\int_{i}^{i+1}f$ is a bounded increasing sequence, hence convergent.

Now, how would you show that if $\sum_{n=1}^{\infty}\int_{n}^{n+1}f$ converges then $\int_{1}^{\infty}f$ converges? The only thing I can see is that if $c\in\mathbb{R}$ with $c>1$ then $\int_{1}^cf\le \sum_{n=1}^{\infty}\int_{n}^{n+1}f$. But does it imply that $\lim_{c\to\infty}\int_{1}^cf$ exists? (I'm not sure, because this is the limit of a function, not just a sequence).

Thank you.

• Yes. There exists a monotone convergence theorem for limits of functions just the same as the one for sequences. – Wojowu May 9 '16 at 20:44
• @Wojowu I see. Does it also converge to the supremum of the function? – Talexius May 9 '16 at 20:55
• If the function is increasing - yes. If it's decreasing, then it converges to its infimum. – Wojowu May 9 '16 at 20:55
• Oh, right, thank you @Wojowu – Talexius May 9 '16 at 21:00

The $n$th partial sum of the series is $\int_1^{n+1}fdx$ and $\int_1^{\infty}fdx$ is defined as the limit of $\int_1^y fdx$ as $y\to \infty$, now use positivity of $f$.