For a measurable function $f$ on $[1,+\infty)$, which is bounded on bounded sets, define $a_n=\int\limits_n^{n+1}f dm$ for each natural number $n$. Is is true that $ f $ is integrable over $[1,+\infty)$ if and only if the series $\sum\limits_{n=1}^{\infty}a_n$ converges absolutely?
I have done like below:
We have $\sum\limits_{n=1}^{\infty}|a_n|=\sum\limits_{n=1}^{\infty}|\int\limits_n^{n+1}f \,dm|\leq \sum\limits_{n=1}^{\infty} \int\limits_{[n,n+1)}|f|\,dm=\sum\limits_{n=1}^{\infty}\int\limits_{[1,+\infty)}|f|\chi_{[n,n+1)}\,dm=\int\limits_{[1,+\infty)}\sum\limits_{n=1}^{\infty}|f|\chi_{[n,n+1)}=\int\limits_{[1,+\infty)}|f|\,dm$. If $f$ is integrable, then $\sum\limits_{n=1}^{\infty}|a_n|<+\infty$ and so $\sum\limits_{n=1}^{\infty}a_n$ converges absolutely.
But is the converse true? I have no answer. Would anyone please help me?
Best Answer
Consider $f(x)=\sin(2\pi x)$. Is it integrable over $[1,\infty)$? What are the $a_n$?
For the other direction, consider $g(x)=\sin(\pi x)/x$. It is integrable over $[1,\infty)$? What is $\sum|a_n|$?