# “Rules of thumb” to decide which convergence test is most appropriate

Are there some general "rules of thumb" that people, who lack enough experience or intuition, could apply to decice which convergence test (such as the ratio test, the nth root test, or Raabe's test etc) is most appropriate to test the absolute convergence (or find the radius of convergence) of a series

$$\sum_{n=1}^{\infty} = \left|a_n\right|$$

just from looking at the definition of its elements $a_n$?

If such general rules exist, what are the underlying (deeper mathematical) motivations for them?

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Is there a typo? – timur Sep 4 '12 at 22:43
One good rule of thumb is to apply the "easiest" tests first -- so if it looks easy to take the ratio of consecutive terms, take ratios. (This includes if $a_n$ contains factorials, since most of the factors will cancel.) If $a_n$ contains $n$th powers, try the root test first. – Michael Lugo Sep 4 '12 at 22:47

Basically just what Michael Lugo said: do the easiest test first, then the second easiest, the third easiest and so on. I was taught [in undergraduate physics (credit to my lecturer)] to do these three tests in the order laid out:

## The ratio test for convergence (d'Alembert test):

Assume $a_{n} > 0$ for all $n$ . Let $R = \lim_{n \to \infty} \frac{a_{n+1}}{a_{n}}$.

If $R < 1$, the series $\sum_{n=n_{0}}^{\infty} a_{n}$ converges.

If $R > 1$, the series $\sum_{n=n_{0}}^{\infty} a_{n}$ does not converge.

If $R = 1$, the series $\sum_{n=n_{0}}^{\infty} a_{n}$ may or may not converge.

## The root test for convergence (Cauchy test):

Assume $a_{n} > 0$ for all $n$ . Let $Q = \lim_{n \to \infty}$ $(a_{n})^{1/n}$.

If $Q < 1$, the series $\sum_{n=n_{0}}^{\infty} a_{n}$ converges.

If $Q > 1$, the series $\sum_{n=n_{0}}^{\infty} a_{n}$ does not converge.

If $Q = 1$, the series $\sum_{n=n_{0}}^{\infty} a_{n}$ may or may not converge.

## The integral test for convergence:

Assume that $f(x)$ is a monotonically decreasing and positive function on the interval $N \le$ x < $\infty$, and let $a_{n} = f(n)$ for integer $n$.

If $\int_{N}^{\infty} f(x) dx$ is finite, the series $\sum_{n=n_{0}}^{\infty} a_{n}$ converges.

If $\int_{N}^{\infty} f(x) dx$ is infinite, the series $\sum_{n=n_{0}}^{\infty} a_{n}$ does not converges.

Also, sometimes if you have an easy looking sum, you can tell by inspection.

Another note: always always take the limit to infinity; testing the first $10^{10^{10}}$ terms, you haven't proved that it converges.

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Hi thanks for this answer, it looks very helpful. I did not know the integral test ... – Dilaton Sep 12 '12 at 10:30
Although performing the integral test is a long-winded process, it always gives you a definite answer and so it is a powerful tool (although, are Maple or Mathematica xD). – User 17670 Sep 12 '12 at 11:02
If the answer was helpful, please consider marking it as accepted, this will give more interest in the question (rep for you) and more interest in the answer (rep for me). – User 17670 Sep 12 '12 at 12:49
Sure, dont worry ;-) I always accept the to me most useful answers. Sometimes it takes some time because I have to see if I can reproduce suggested calculations for example etc, but yours here is obviously very nice and the integral test I did not know before is very intuitive too. Cheers – Dilaton Sep 12 '12 at 13:03