# For which values of x does the sequence $a_n$ converge? (ratio test for $a_n=\frac{x^n}{n^3}$)

I was wondering if anyone could offer any tips or hints as to how to tackle this question, I've tried looking around on here and in my notes but I can't seem to find anything.

Consider the sequence

$${a_n=\frac{x^n}{n^3}}.$$

B1. Apply the ratio test to ﬁnd out for which values of $$x$$ the sequence $$(a_n)$$ converges.

B2. For which value(s) of $$x$$ does the ratio test provide no information?

B3. Does $$(a_n)$$ converge or diverge for these values? Justify your answers.

B4. Give an example of a sequence that converges if $$-1 and diverges otherwise. Justify your answer.

• In B$, you ask [...] converges if$1<x<1$[...]. Do you mean$-1<x<1$? – Alain Remillard Oct 31 '19 at 15:32 • Ah yes I do sorry – CAHL632 Oct 31 '19 at 16:55 ## 1 Answer By ratio test we obtain $$\left|\frac{x^{n+1}}{(n+1)^3}\frac{n^3}{x^n}\right|=|x|\left(\frac n{n+1}\right)^3 \to |x|$$ then for convergence we need $$|x|<1$$, then study separately the cases $$|x|=1$$. • Thank you for this, do you have any suggestions on tackling B4 or an example I could use? – CAHL632 Nov 1 '19 at 19:06 • @CAHL632 We have that$\frac{\left(\frac12\right)^n}{n^3}$converges and$\frac{2^n}{n^3}\$ diverges. – user Nov 1 '19 at 21:04
• I think the question is looking for just one sequence which adheres to both rules instead of an example of each. – CAHL632 Nov 2 '19 at 11:45