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Ratio test for series


One of the most practical convergence tests for series is the ratio test. Just a limit of a ratio defined by the general term of the series can tell us if the series converges or diverges.

We mention that in the French mathematical school this test is called the d’Alembert rule. This is because it was founded by the French mathematician d’Alembert.

What is a ratio test?

Convergent series are very important in some mathematical models. It is also useful for developing a regular function around a point.

We recall that a series $\sum_{n= 0}^{+\infty}u_n$ is convergent if the partial sums sequence $(S_n)_n$ defined by $S_n=u_0+cdots+u_n$ has a limit. It is called divergent if the limit is $\pm\infty$ of the limit does not exist at all.

Here, in this paragraph, we introduce a useful test for the convergence of series. This is called the ratio test for series and is summarized in the following result.

Theorem: Assume a sequence of real numbers $(u_n)_n$ satisfies $u_n>0$ and there exists a real number $\ell$ such that $$ \lim_{n\to+\infty} \frac{u_{n+1}}{u_n}=\ell.$$ Then the following assertions hold:

  • The series $\sum_{n=0}^{+\infty}u_n$ is convergent if $\ell<1$;
  • it is divergent if $\ell>1$;
  • finally, if $\ell=1$ we cannot conclude.

An overview of the bounded ratio test

D’Alembert was a French mathematician and philosopher, born in Paris in 1717 and died in 1783. He had a successful career at law school. However, the legal profession did not appeal to d’Alembert, so he decided to take courses in medicine. Again, only after a while does he turn to mathematics. He later became one of the greatest masters of mathematics.

Examples of application of the convergent test

We give some applications of the ration test. Here we give classical series.

Example 1: Discuss the convergence of the following series \begin{align*} \sum_{n=0}^{+\infty} \frac{1}{n!},\quad \sum_{n=0}^{+\infty} \frac{n}{2^n}.\end{align*} Solution: Let us put $u_n=\frac{1}{n!}$. Then we have \begin{align*} \frac{u_{n+1}}{u_n}=\frac{\frac{1}{(1+n)!}}{\frac{1}{n!}}=\frac{1}{n+1}.\end{align*} Clearly the limit of this ratio is $0$. Then the series $\sum_{n=0}^{+\infty} \frac{1}{n!}$ is convergent. On the other hand, select $v_n=\frac{n}{2^n}$. So that $$ \frac{v_{n+1}}{v_n}=\frac{n+1}{2^{n+1}}\times \frac{2^n}{n}=2 \frac{n+1}{n}.$$ The limit of this ratio is $2$. Thus the ratio test, the series $ \sum_{n=0}^{+\infty} \frac{n}{2^n}$ is divergent.

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