# Sequence proof(ratio test)

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.

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

Let $$s=(1+r)/2.$$
If $$0\le r< 1$$ then $$0 so for all but finitely many $$n$$ we have $$(a_n)^{1/n} and hence $$0\le a_n so $$\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.