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Just wondering how I would formally prove the root test for series which is the following,

$$\sum_{n=1}^{\infty}a_n, a_n \ge0 $$ for $$n\ge N$$ The root test states that if ${a_n} $ satisfies,

$$\lim_{n \to\infty}\sqrt[n]{a_n}=\rho$$ The series $\sum_{n=1}^{\infty}a_n$ converges if $\rho<1$ and diverges if $\rho >1$, and is inconclusive if $\rho=1$. I'm not really sure how to begin, first steps or complete solutions welcome.

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  • $\begingroup$ The idea is to compare $a_n$ to a geometric series. $\endgroup$ – mookid Feb 28 '14 at 19:43
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If $\rho>1$ then you know that for $n\geq K$ you have $\sqrt[n]{a_n}>1+\epsilon \Longleftrightarrow a_n>(1+\epsilon)^n$ for some $\epsilon>0$, so the series is bounded from below by a divergent geometric series.

Similarly if $\rho<1$ you know that for $n\geq K$ you have $\sqrt[n]{a_n}<1-\epsilon \Longleftrightarrow a_n<(1-\epsilon)^n$ for some $\epsilon>0$, so the series is bounded from above by a convergent geometric series.

To see that the test is inconclusive when $\rho=1$, consider the following examples:

  • $\sum\limits_{n=1}^{\infty}\frac1n$ diverges and the test gives $\sqrt[n]{\frac1n}=\frac1{\sqrt[n]n} \rightarrow 1$
  • $\sum\limits_{n=1}^{\infty}\frac1{n^2}$ converges and the test gives $\sqrt[n]{\frac1{n^2}}=\frac1{(\sqrt[n]n)^2} \rightarrow 1$

Remarks:

$\sqrt[n]n\rightarrow 1$ follows from $n < (1+\epsilon)^n$ for sufficiently high $n$ and $\epsilon>0$, because then $\sqrt[n]{n}<\sqrt[n]{(1+\epsilon)^n}=1+\epsilon$.

This test can be modified to test the absolute convergence of a series by computing $\varlimsup_{n\to\infty}\sqrt[n]{|a_n|}$, the only thing to note is that the convergence of the sum fails by the basic $a_n\rightarrow 0$ test, not by bounding below by a geometric series.

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  • $\begingroup$ Your first paragraph only establishes that $a_n$ does not converge absolutely. (Similarly your second paragraph proves that $a_n$ does converge absolutely, but there it's not a problem). $\endgroup$ – Jack M Feb 28 '14 at 19:54
  • $\begingroup$ @Jack M I thought the OP states $a_n \geq 0$ $\endgroup$ – user2345215 Feb 28 '14 at 19:58
  • $\begingroup$ @user2345215 Still, it might be better to answer the general situation, when $\limsup\limits_{n\to\infty} \sqrt[n]{\lvert a_n\rvert} > 1$, the series diverges because the terms don't converge to $0$. $\endgroup$ – Daniel Fischer Feb 28 '14 at 20:05
  • $\begingroup$ But what is $K$? $\endgroup$ – Awn Nov 30 '17 at 22:05

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