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Can anyone help me solve this question? I was trying to verify that it was true first by testing values of $r$ and it appeared no matter what values I used it seemed to approach 1 at infinity.

Consider a sequence $(a_n)_{n=1}^\infty$ such that $a_n \geqslant 0$ for all $n \in \mathbb{N}$ and $\lim_{n \rightarrow\infty} \sqrt[n]{a_n} = r$ for $r \in \mathbb{R}$. Prove that the series $\sum_{n=1}^\infty a_n$ converges if $r < 1$ and diverges if $r > 1$.

This result is known as the Root Test.

Hint: the proof is similar to that of the Ratio Test.

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    $\begingroup$ Read the proof given in a text book. We can help you if you get stuck. $\endgroup$ May 29 '19 at 23:54
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    $\begingroup$ This seems like a homework problem. Could clarify where you are stuck or struggling? What have you tried so far? $\endgroup$ May 30 '19 at 0:05
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    $\begingroup$ I think you mean $\sqrt[n]{a_n}$ instead of $\sqrt{n}a_n$ $\endgroup$
    – saulspatz
    May 30 '19 at 0:27
  • $\begingroup$ yes I do, someone edited it $\endgroup$ May 30 '19 at 0:28
  • $\begingroup$ Do you think you could fix it? it won't allow me to edit it $\endgroup$ May 30 '19 at 0:32
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Let $s=(1+r)/2.$

If $0\le r< 1$ then $0<r<s<1,$ so for all but finitely many $n$ we have $(a_n)^{1/n}<s<1 ,$ and hence $0\le a_n<s^n,$ so $\sum_na_n$ converges by comparison to the geometric series $\sum_ns^n.$

If $r>1$ then $r>s>1,$ so for all but finitely many $n$ we have $(a_n)^{1/n}>s,$ and hence $a_n>s^n>1,$ so the rest is (I hope) obvious.

If $r=1$ then $\sum_na_n$ may or may not converge. For example $(1/n)^{1/n}$ and $(1/n^2)^{1/n}$ both $\to 1$ as $n\to \infty$ but $\sum_n (1/n)$ diverges and $\sum_n(1/n^2)$ converges.

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