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So i proved that $$\lim_{n\to \infty} U(n+1) - U(n) = 0$$ and we also have $a< U(n) < b$ What can we say?

Does this prove that the serie converge?

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  • $\begingroup$ Is $U(n)$ a sequence of numbers? $\endgroup$ – user99914 Nov 7 '14 at 22:33
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Counter-example: $\sin(\sum_{i=1}^{n}{\frac{1}{i}})$.

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Consider the series $$1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}- \frac{1}{6}-\frac{1}{6}-\cdots.$$

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