# Root Test: What happens if $\lim\limits_{n\rightarrow \infty} \sqrt[n]{|a_n|} =1$?

Root Test: For a series $\sum\limits_{n=1}^\infty a_n$, let $\ell = \lim\limits_{n\rightarrow \infty} \sqrt[n]{|a_n|}$. Series is convergent if $\ell <1$ and divergent if $\ell \geq 1$. I have the question about $\ell =1$. Some of the standard calculus book mention that the root test fails for $\ell =1$ and some other mention it is divergent in this case. Your thought please.

My thought: $\lim\limits_{n\rightarrow \infty} \sqrt[n]{|a_n|} \geq 1$ imples that $|a_n| \geq 1$ for sufficiently large $n$ which means $\lim\limits_{n\rightarrow \infty} |a_n|\geq 1 \neq 0$. Therefore divergent by the Divergence test.

• You can't say anything. It can converge or diverge ! For example $\sum_{n}1^n$ diverge, but $\sum_{n}\frac{1}{n^2}$ converge – Surb Jul 11 '16 at 19:02
• Please, provide examples. – Mr. MBB Jul 11 '16 at 19:02
• I gave you example :) – Surb Jul 11 '16 at 19:03
• No text says $l=1$ implies divergence. – zhw. Jul 11 '16 at 19:17
• @zhw: Look at the book Introduction to Analysis, by Edward Gaughan, page no. 185. – Mr. MBB Jul 11 '16 at 19:29

Hint. One may observe that $\displaystyle\sum_{n\geq1} \frac{1}{n}$ and $\displaystyle\sum_{n\geq2} \frac{1}{n\log^2 n}$ behave differently, the former is divergent, the latter is convergent, in both cases $$\lim\limits_{n\rightarrow \infty} \sqrt[n]{|a_n|} =1.$$
• It's easier to look at $\sum 1/n^2$ in place of your second series. – zhw. Jul 11 '16 at 19:19
• @Mr.MBB. Just apply en.wikipedia.org/wiki/Cauchy_condensation_test. and you obtain $\sum\frac{1}{n^2}$, which converges – b00n heT Jul 11 '16 at 20:13
$a_n=\frac{1}{n}$ then the serie diverge but $\sqrt[n]{a_n}$ converge to $1$
$a_n=\frac{1}{n^2}$ then the serie converge but $\sqrt[n]{a_n}$ converge to $1$