How to Prove the Root Test For Series 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.
 A: 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.
