For a measurable function, $f$, on $[1, \infty)$ which is bounded on bounded sets, define $a_n = \int_n^{n+1} f$ for each natural number $n$. Is it true that $f$ is integrable over $[1, \infty)$ if and only if the series $\sum_{n=1}^\infty a_n$ converges? Is it true that $f$ is integrable over $[1, \infty)$ if and only if the series $\sum_{n=1}^\infty a_n$ converges absolutely?
Here is what I am thinking:
Let $\sum_1^\infty a_n$ be convergent. $\sum_1^\infty a_n$ = $\sum_1^\infty (\int_n^{n+1}f)$. Since n are natural numbers, $n
Conversely, let f be integrable over $[1, \infty)$.
Given the above is true. To show for absolute convergence we would need to show for $\sum_1^\infty |a_n|$.
Am I headed in the right direction?