TITLE: When is infinite sum bounded by an integral?
QUESTION [8 upvotes]: What conditions does $f(i)$ have to have for the inequality $\sum_{i=1}^{\infty} f(i) \leq \int_{i=0}^{\infty} f(i)$ to hold?  
My guess is that $f(i)$ has to be monotone and that the sum and integral should both exist. Is it right that you don't need to worry about whether $f$ is increasing or decreasing for example?

REPLY [8 votes]: You're pretty much right. If $f(x)$ is a decreasing integrable function, then $f(i) \le \int_{i-1}^i f(x)\,dx$. By summing several such inequalities, we deduce that $\sum_{i=1}^N f(i) \le \int_0^N f(x)\,dx$. Then take the limit of both sides as $N\to\infty$; assuming both limits exist, we conclude that $\sum_{i=1}^\infty f(i) \le \int_0^\infty f(x)\,dx$.
(In fact, if $f$ is nonnegative, as presumably you want to assume, then you only need to show that $\int_0^\infty f(x)\,dx$ exists; if so, then the sequence $\big\{ \sum_{i=1}^N f(i) \big\}$ is an increasing sequence bounded above, hence its limit automatically exists.)
If $f$ is increasing, then all the inequalities (starting with $f(i) \ge \int_{i-1}^i f(x)\,dx$) will be reversed.