An unusual issue with the ratio test

The ratio test is a very useful test for convergence, whether of sequences or of series. As usual, it comes in many forms: here is one.

Note here that we are using the standard notions of convergence and divergence for sequences and series of real (or complex) numbers here.

The ratio test for sequences and series

Let (x_n)_{n=1}^\infty be a sequence of non-zero real (or complex) numbers, and suppose that |x_{n+1}|/|x_n| \to L as n \to \infty. (Here you may allow L=\infty if you wish.)

  • If L<1, then the series \sum_{n=1}^\infty x_n is absolutely convergent, and (hence) x_n \to 0 as n \to \infty.
  • If L>1, then the series \sum_{n=1}^\infty x_n is divergent. Moreover, |x_n| \to \infty as n \to \infty, and (hence) the sequence (x_n) is divergent.
  • If L=1 then the ratio test is inconclusive, and you need to try another test.

In my FAQ document for Mathematical Analysis, I recommend using the ratio test if your terms involve things like powers of constants (e.g. 2^n) or factorials. For example, the ratio test quickly shows you the divergence of the  sequence (x_n) when the terms x_n are 3^n/2^n or n!/(1000)^n, and the convergence to zero of the sequence (x_n) when x_n = (1000)^n/n!.

If, instead, the terms are simply rational functions of n, then the ratio test is always inconclusive! For example, the ratio test does not help you when x_n  is n, or (2n^2+1)/n^3, or (3n+2)/n^3. For all three of these examples, the limiting ratio L exists, but is equal to 1

Of course, the limiting ratio L need not exist at all. (You can still make further progress using \limsup and \liminf instead.) So that is another obvious way for the ratio test, as stated above, to be inconclusive.

But one rather unusual way for this version of the ratio test to be inconclusive is, I would argue, the following case.

Suppose that x_n=2^{-2^n}. Common sense tells us that this sequence converges to zero very fast. This is, indeed, true! But what happens if we try to use the ratio test in the form we have stated it above to prove that the sequence converges to 0?

Well, we have,

|x_{n+1}|/|x_n| = 2^{-2^{n+1}}/2^{-2^n} = 2^{-2^n}=x_n\,.

Is this a problem? Well, if you are trying to investigate the limiting ratio, you now find yourself back where you started! That is, you have a sequence which you think converges to zero, and you think you might prove it using the ratio test …

Of course, if you use a suitable alternative version of the ratio test, the fact that the ratios are always less than 1/2 is immediately conclusive.

Note that the same ratio test issue arises if you take x_n = 2^{2^n}.

Joel

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