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I'm trying to show that if lim $(\sqrt{n} |a_n|) = L, L >0$, then $\sum a_n$ cannot converge absolutely. I'm trying to work it from the definition of a limit that for any $\epsilon > 0$, there exists an $n$ such that $|\sqrt{n}|a_n| - L| < \epsilon$, but I'm not really sure how to use this. What technique should I be using to prove this?

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There's a general theorem which states: If $a_n \geq 0$, $b_n > 0$ and $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = L > 0$, then $\sum a_n$ converges if and only if $\sum b_n$ converges. (You should try to prove this).

In this particular setting, you can use the sequences $|a_n|$ and $1/\sqrt{n}$. Then:

$$\lim_{n \rightarrow \infty} \frac{|a_n|}{1/\sqrt{n}} = \lim_{n \rightarrow \infty} \sqrt{n}|a_n| = L > 0$$

Then since $\sum 1/\sqrt{n}$ diverges, so does $\sum |a_n|$

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Your definition of limit isn't quite right: it should be "for any $\epsilon>0$ there exists an $m$ such that $|\sqrt n|a_n|-L|<\epsilon$ for every $n\geq m$.

Hint: use this to show that if $\lim(\sqrt n|a_n|)=L$ then there exists $m$ with $|a_n|>\frac{L}{2\sqrt n}$ for $n\geq m$, then use the fact that $\sum\frac 1{\sqrt n}$ doesn't converge.

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we are given the condition that $\lim_{n\rightarrow\infty}\sqrt{n}|{a_n}|=L>0$. then for $\epsilon=\frac{L}{2}$, there exists an $N$ such that $n\geq N$ implies $|\sqrt{n}|a_n|-L|<\frac{L}{2}$.

Then for $n\geq N$, $|a_n|\geq\frac{L}{2}\cdot\frac{1}{\sqrt{n}}---(*)$.

Therefore if we sum over $n$ greater than $N$, the RHS of the (*) diverges. So $\sum a_n$ does not converges absolutely.

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