Once I take my integral of f'(x) from n+1 to n I'll get f(n+1)-f(n). Since I already needed a function that's limit diverged, shouldn't my sum also diverge? This leads me to believe that this is a trick question and is impossible.
2026-03-26 13:30:28.1774531828
Is there a function for which the sum converges while the limit diverges?
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The upper expression simplifies to $\lim_{n \to \infty} f(n)$.
So take $f(x) = \sin(\pi x)$. $f(n) \equiv 0$ but $f(x)$ oscillates, so it is divergent.