So, the ratio test says that when the limit is bigger than one it does not converge. Can you explain why it can't never ever be bigger than one?

And what to do when limit =1?

  • $\begingroup$ The ratio test should work for geometric series. $\endgroup$ – Joel Cohen Mar 28 '13 at 18:04
  • $\begingroup$ @JoelCohen soo? $\endgroup$ – MITjanitor Mar 28 '13 at 18:05
  • $\begingroup$ yeah??? Now, the question was, 'why' does it actually 'work'? Why does something not converge when the result is '2' for example? $\endgroup$ – user1095332 Mar 28 '13 at 18:08
  • $\begingroup$ What do you mean by 'the limit'? Presumably you mean the limit as $n$ tends towards infinity of $\frac{a_{n+1}}{a_n}$. $\endgroup$ – Tara B Mar 28 '13 at 18:10
  • $\begingroup$ I see that you edited your question and added "And what to do when limit =1?" In that case you need better analysis. The test cannot in that case give any conclusions, as far as I know. $\endgroup$ – user67878 Mar 28 '13 at 19:06

In order for $\sum_n a_n$ to converge you need $a_n \to 0$; but if $\left|\frac{a_{n+1}}{a_n}\right|$ tends to some limit $\ell > 1$ then this can't happen.

  • $\begingroup$ Why does the limit for x goes to infinity doesn't need to be exactly 0? $\endgroup$ – user1095332 Mar 28 '13 at 18:52
  • $\begingroup$ Intuitively it's because if the ratio tends to a limit greater than $1$ then the sequence is 'eventually increasing'. Technically... well, work through the details. $\endgroup$ – Clive Newstead Mar 28 '13 at 18:59

The ratio test, intuitively, measures how far or close your sequence is from a geometric one. For instance, say your sequence $\langle a_n\rangle$ with $a_n>0$ is such that $$\lim \frac{a_{n+1}}{a_n}=\ell < 1$$

Now, choose $r$ with $\ell<r<1$. Then, for some sufficiently large $N$ and $n\geq N$, we have that $$\frac{a_{n+1}}{a_n}<r$$

so $${a_{n+1}}<r{a_n}$$

This means that for $i=0,1,2,\dots$ and any $n\geq N$


It follows that the "tail" $$\sum_{n=N+1 }^\infty a_{n}$$

is bounded above by a geometric series, which is convergent. Thus, the tail converges and the whole sequence will do.

Similarily, if $\lim\frac{a_{n+1}}{a_n}=\ell >1$, your series will grow faster than a geometric series of with ratio $r>1$, which we know is divergent, so your series will also diverge.

The moral of the story is that anything that behaves better than a convergent geometric series will converge, and anything that behaves worse than a divergent geometric series will diverge.


You know that the geometric series $\sum_{n \ge 0} z^n$ converges if and only if $|z| < 1$. So the idea of the ratio test is to compare your sequence $u_n$ to a geometric one by computing the ratio $\frac{u_n}{u_{n+1}}$ (which is constant if and only if $u_n$ is a geometric sequence).

If the ratio converges to some limit $l < 1$, $u_n$ is bounded above (for $n$ sufficiently large) by a geometric series of ratio $r \in ]l,1[$, which converges. If the ratio converges to some limit $l > 1$, $u_n$ is bounded below (for $n$ sufficiently large) by a geometric series of ratio $r \in ]1,l[$, which diverges.


This is only an answer of the added question in the edited version of your question, of the case when limit is equal to $1$. The case when the limit is strictly less than one or strictly grater than one is explained in a nice way by other users. In the case when the limit is equal to $1$ the ratio test is inconclusive, see, for instance, these two well-known examples:

1) $\lim_{n\to\infty} \dfrac {\frac {1}{(n+1)^2}}{\frac {1}{n^2}}=1$ and the series $\sum_n \dfrac {1}{n^2}$ is convergent.

2) $\lim_{n\to\infty} \dfrac {\frac {1}{n+1}}{\frac {1}{n}}=1$ and the series $\sum_n \dfrac {1}{n}$ is divergent.

So some series will diverge and some will converge if the limit is equal to $1$.


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