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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.

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    $\begingroup$ 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 $\endgroup$
    – Surb
    Jul 11, 2016 at 19:02
  • $\begingroup$ Please, provide examples. $\endgroup$
    – Mr. MBB
    Jul 11, 2016 at 19:02
  • $\begingroup$ I gave you example :) $\endgroup$
    – Surb
    Jul 11, 2016 at 19:03
  • $\begingroup$ No text says $l=1$ implies divergence. $\endgroup$
    – zhw.
    Jul 11, 2016 at 19:17
  • $\begingroup$ @zhw: Look at the book Introduction to Analysis, by Edward Gaughan, page no. 185. $\endgroup$
    – Mr. MBB
    Jul 11, 2016 at 19:29

2 Answers 2

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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 $$ \lim\limits_{n\rightarrow \infty} \sqrt[n]{|a_n|} =1. $$

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  • $\begingroup$ Please prove a proof of your second example. $\endgroup$
    – Mr. MBB
    Jul 11, 2016 at 19:10
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    $\begingroup$ It's easier to look at $\sum 1/n^2$ in place of your second series. $\endgroup$
    – zhw.
    Jul 11, 2016 at 19:19
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    $\begingroup$ @Mr.MBB. Just apply en.wikipedia.org/wiki/Cauchy_condensation_test. and you obtain $\sum\frac{1}{n^2}$, which converges $\endgroup$
    – b00n heT
    Jul 11, 2016 at 20:13
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If $a_n=\frac{1}{n}$ then the series diverges but $\sqrt[n]{a_n}$ converges to $1$.

If $a_n=\frac{1}{n^2}$ then the series converges but $\sqrt[n]{a_n}$ converges to $1$.

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