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I just found out about convergence tests , found them on Wikipedia and I've got this question : Are you allowed to use , for example , the ratio test if your sequence is NOT defined as a sum ? Let's say $a_n=1/(n!)$. Are you allowed to apply the ratio convergence test on this sequence ?

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    $\begingroup$ Yes, there is a ratio test for sequences too. See for example here. $\endgroup$ – ah11950 Apr 20 '14 at 15:08
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That depends on the convergence test. In general, just because a test doesn not indicate that a series does converge, that doesn't mean that it cannot converge.

For the ration test in particular, you have that $\sum_{k=1}^\infty a_k$

  • converges absolutely if $\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| < 1$,
  • diverges if $\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| > 1$,
  • can do either if $\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| = 1$

For $\sum_{k=1}^\infty \frac{1}{k}$, i.e. $a_k = \frac{1}{k}$, you have $\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| = \lim_{k\to\infty} \left|\frac{k}{k+1}\right| = 1$, so the ratio test doesn't provide you with any information about $\sum_{k=1}^\infty \frac{1}{k}$

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    $\begingroup$ I was under the impression that the OP was asking about convergence tests for series translating to convergence tests for sequences... $\endgroup$ – ah11950 Apr 20 '14 at 15:12
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    $\begingroup$ @ah11950 Ah, OK. I interpreted "if your sequence is NOT defined as a sum" that for some sequence $a_k$, the sum $\sum a_k$ diverges, i.e. is not defined. Since the answer is not wrong, but merely fails to answer the actual question, i'm going to let it stick around, I think. $\endgroup$ – fgp Apr 20 '14 at 15:31

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