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Consider $\sum_{n=1}^{\infty} \frac{1}{n^{3}}$. We know that it converges.

Given $k\in\left(0,\infty\right)$. Is it then "okay" to say that there exists a $j\in\mathbb{N}$ such that $\sum_{n=j}^{\infty} \frac{1}{n^{3}}<k$? Or does it need some extra argumentation?

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up vote 2 down vote accepted

It always depends on the audience, but for an audience knowledgeable about sequences and series this should be obvious enough not to require further detailed arguments.

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I agree since it is an elementary fact that a series converges only if its terms tend to $0$. –  robjohn Sep 26 '12 at 18:02
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The fact $\sum\limits_{n=1}^{\infty} \frac{1}{n^{3}}$ that converges means that converges the sequence of partial sums $S_N=\sum\limits _{n=1}^{N} \frac{1}{n^{3}}$

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