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Mathematics
Calculus
Understanding the nth-term test for Divergence
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[QUOTE="chwala, post: 6845123, member: 287397"] [B]TL;DR Summary:[/B] I am looking at the theorem; It states that 'If the sequence ##a_n## does not converge to 0, then the series ##\sum_{n=0}^\infty a_n## diverges. There are several examples that i have looked at which are quite clear and straightforward, e.g ##\sum_{n=0}^\infty 2^n## it follows that ##\lim_{n \rightarrow \infty} {2^n}=∞## thus going with the theorem, the series diverges. Now let's look at the example below; ##\sum_{n=1}^\infty \dfrac{1}{n}## it follows that, ##\lim_{n \rightarrow \infty} {\left[\dfrac{1}{n}\right]}=0## the text indicates that the nth term test for divergence does not apply...infact we know that the series diverges (using p -series)... now back to my question. What is the relevance of this theorem, if any ...why use 'does not converge to 0, if converging to 0 after all has no implication' whatsoever? the theorem is not complete by itself in my thinking. Can we say converging to 0, may or may not mean convergence? implying that we have to use other tools to ascertain that? I am also looking also at the nth-term test of a convergent series- Now if a series ##b_n## diverges, does the sequence ##b_n## converge to 0 or not? just going with what is stated in the theorems. thanks. [/QUOTE]
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Calculus
Understanding the nth-term test for Divergence
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