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 March 2nd, 2009, 12:10 PM #1 Newbie   Joined: Mar 2009 Posts: 8 Thanks: 0 Confused I am given a_n > 0, lim (n*a_n) = l, l != 0... show that \Sum (a_n) is divergent. I would like a hint where to start. I thought that since the limit n*a_n = l, that means that n*a_n is bounded by some value greater than 0... but I am unsure if that can help me show that a_n is unbounded. Or am I completely in the wrong area here. Thank you
 March 2nd, 2009, 01:35 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Confused If $a_n>0,$ and $\lim_{n\mapsto\infty}na_n=l\neq0,$ then $\lim_{n\mapsto\infty}\frac{a_n}{\frac1n}=l,$ so by the comparison of limits test, $\sum_{n=0}^\infty a_n$ converges iff $\sum_{n=0}^\infty\frac1n$ converges.
 March 2nd, 2009, 03:52 PM #3 Newbie   Joined: Mar 2009 Posts: 8 Thanks: 0 Re: Confused Thanks... The limit comparison test isn't a theorem we can use, but it gave me an idea on how to do the problem if a_n >0 for all and and lim (n*a_n) = l, that implies that n*a_n is bounded. Thus there exist real numbers c,d such that c <= n*a_n <= d <=> 1/n * c <= a_n. Since Sum (1/n) diverges (sum c(1/n) diverges) then by the comparison test Sum(a_n) must diverge. by similar logic if lim n^2*a_n, converges, a_n > 0 then Sum (a_n) must converge.

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