When is infinite sum bounded by an integral?

Solution 1:

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.