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I've been having troubles finding information about the root test for sequences (not series!) that someone mentioned to me (I was interested in seeing a proof, at least). I've looked for it in many books and online, but could only find information about a root test for series. I have attached an image containing the test - is this genuine or is that person wrong?

Root test for sequences convergence?

Root test (D'Alembert test): May $(a_n)_{n\ge0}$ be a sequence with strictly positive terms. If $l\in\mathbb R\cup\{-\infty,+\infty\}$ exists so that $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}=l$ then $\lim\limits_{n\to\infty} \sqrt[n]{a_n}=l$.

Martin Sleziak
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0ana
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  • It's true. See these notes of Pete L. Clark for a proof. – David Mitra Feb 02 '14 at 16:52
  • By the way, note it's not giving a test for the convergence of $(a_n)$; it's just saying that if the limit of the ratios exists, then so does the limit of the roots and in this case the two limits are the same. – David Mitra Feb 02 '14 at 16:53
  • Thank you for your answer, David, but aren't the notes about series? I'm looking for information about the convergence of sequences (not series), I will edit my question to make that clearer. – 0ana Feb 02 '14 at 16:57
  • See Lemma 3 on page 3 (it applies to any sequence of positive terms). – David Mitra Feb 02 '14 at 16:58
  • I answered too soon, I should have read more carefully! Thank you so much, this answeres my question. Would you post it as an answer for me to be able to accept it? – 0ana Feb 02 '14 at 17:00

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It's true. See these notes (Wayback Machine) of Pete L. Clark for a proof (Lemma 3 on page 3). Another proof can be found here at MSE.

(By the way, note the result is not giving a test for the convergence of$(a_n)$; it's just saying that if the limit of the ratios exists, then so does the limit of the roots and in this case the two limits are the same. )

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David Mitra
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