Teste de raízes

In the previous section, we talked about a way to tell when a series converges. Another very useful test is called the root test, which states the following:

Theorem: For the series $\sum a_n$, consider the number $$L=\lim_{n\to\infty} |a_n|^{\frac{1}{n}}=\lim_{n\to\infty} \sqrt[n]{|a_n|}.$$ Then we have the following:

1. If $L<1$, then $\sum a_n$ converges absolutely.

2. If $L>1$, the series diverges.

3. If $L=1$, then the test is inconclusive.

Let's look at an example: Suppose we want to see whether the series $$\sum_{n=1}^{\infty} \frac{3^{3n+2}}{n^n}$$ converges. Well, observe that $$L=\lim_{n\to\infty} \bigg|\frac{3^{3n+2}}{n^n}\bigg|^{\frac{1}{n}}=\lim_{n\to\infty} \frac{3^{3+\frac{2}{n}}}{n}=\frac{3^3}{\infty}=0<1.$$ So, by the root test, we conclude that this series converges.

Let's try one more problem together: does the series $$\sum_{n=1}^{\infty} \frac{(-17)^n}{n}$$ converge or diverge? Well first, we have $$L=\lim_{n\to\infty} \bigg|\frac{(-17)^n}{n}\bigg|^{\frac{1}{n}}=\lim_{n\to\infty} \frac{17}{n^{\frac{1}{n}}}=\frac{\lim_{n\to\infty} 17}{\lim_{n\to\infty} n^{\frac{1}{n}}}=\frac{17}{\lim_{n\to\infty} n^{\frac{1}{n}}}.$$ But what is the value of the denominator? To find this limit, we set $$y=n^{\frac{1}{n}}.$$ Then by taking the natural logarithm of both sides, $$\ln{y}=\ln{\left(n^{\frac{1}{n}}\right)}=\frac{\ln{n}}{n}.$$ Now, we find the value of $$\lim_{n\to\infty} \ln{y}=\lim_{n\to\infty} \ln{\left(n^{\frac{1}{n}}\right)}=\lim_{n\to\infty} \frac{\ln{n}}{n}.$$ As $n\to\infty$, both $n$ and $\ln{n}$ approach $\infty$. So, we can apply L'Hopital's rule to see that $$\lim_{n\to\infty} \ln{y}=\lim_{n\to\infty} \frac{\ln{n}}{n}=\lim_{n\to\infty} \frac{(\ln{n})'}{(n)'}=\lim_{n\to\infty} \frac{\frac{1}{n}}{1}=0.$$ Therefore, we have $$\lim_{n\to\infty} n^{\frac{1}{n}}=\lim_{n\to\infty} y=e^0=1.$$ Therefore, returning to our original problem, we have $$L=\frac{17}{\lim_{n\to\infty} n^{\frac{1}{n}}}=17>1.$$ Hence, this series diverges.

Now you try!

Indeed, notice that $$L=\lim_{n\to\infty} \left[\bigg|\left(\frac{-n^3+3n}{5+n^3}\right)^n\bigg|\right]^{\frac{1}{n}}$$ $$=\lim_{n\to\infty} \bigg|\frac{-n^3+3n}{n^3+5}\bigg|=\lim_{n\to\infty} \bigg|\frac{-1+\frac{3}{n^2}}{1+\frac{5}{n^3}}\bigg|=1.$$ Since $L=1$, the root test by itself doesn't tell us anything about the convergence of the series. Darn!

In general, the Root Test only works when something is growing geometrically fast or faster. The Root test is a comparison to geometric series, and cannot determine convergence or divergence unless something is growing/decaying like a geometric series.

Examples of growing/decaying geometrically fast or faster: $3^n,2^{-n},n^n,n!$, etc.

Examples of not growing/decaying geometrically fast or faster: $\log n,n^3,n^{100}$, etc. If your series terms only contain terms like these, the Root test will be inconclusive.

Overview of root test

By comparing to an infinite geometric series, we get a convergence test called the root test. Namely, for the series $\sum a_n$, consider the sequence of $n$th roots: $|a_n|^{\frac{1}{n}} = \sqrt[n]{|a_n|}$. If this "root sequence" converges to a number $L<1$, then the series $\sum a_n$ converges absolutely. If the root sequence converges to a number $L>1$, then the series $\sum a_n$ diverges. Otherwise, if the limit $\lim_{n\to\infty} |a_n|^{\frac{1}{n}}$ either equals $1$ or does not exist, then the test is inconclusive.