I was given an exercise to show that the sequence defined by: $$a_{n+1} = {a_n} + \frac{1}{2{a_n}}, {a_0} > 0$$

prove that: $$\lim_{n \to \infty} {a_n} = +\infty$$

I have proven by induction that the sequence is constantly increasing, and therefore it has a limit which is or a certain number, or +/- infinity.

Using the ratio test, it looks like the

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{{a_n} + \frac{1}{2{a_n}}}{a_n} = \lim_{n \to \infty} 1 + \lim_{n \to \infty} \frac{1}{2{a_n}^2} = 1 + 0 = 1$$

And therefore the ratio test inconclusive.

I've got 2 questions:

1.Am I using the ratio test correctly?

2.If I'm using the ratio test correctly, and it is really inconclusive, any hint how to continue?

Thank you.

  • $\begingroup$ The Ratio Test is for series, not sequences. $\endgroup$ Nov 22, 2015 at 17:14
  • $\begingroup$ I'm not sure about the exact name in English of what I called "ratio test", I think it's D'Alembert test. The test says that if An is a positive sequence $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$. If L>1 the sequence converges to infinity, and if L<1 it converges to 0. $\endgroup$
    – Taru
    Nov 22, 2015 at 17:23
  • $\begingroup$ D'Alembert's Test is another name for the Ratio Test, a test for the convergence of series $\endgroup$ Nov 22, 2015 at 17:25
  • $\begingroup$ I've edited my previous comment if you may please verify that we're talking about the same test. $\endgroup$
    – Taru
    Nov 22, 2015 at 17:26

2 Answers 2


We have: $$ a_{n+1}^2 = a_n^2 + 1 + \frac{1}{4 a_n^2}\geq a_n^2 + 1 $$ hence induction gives $a_n^2\geq n$ and $a_n\geq \sqrt{n}$ is enough to ensure $a_n\to +\infty$.


Hint: Your sequence is increasing. If it is bounded above then it has a finite limit which is a strictly positive number. What can you say about this presumed limit? Otherwise, the sequence is unbounded above.

  • $\begingroup$ I've edited my comment to the question regarding of what I've called the ratio test. May you please check what test I'm talking about in English? $\endgroup$
    – Taru
    Nov 22, 2015 at 17:36

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