show this series diverge

wellfed · #1 $Old$ July 3rd, 2009, 04:42 PM

Hi I am stuck on the following problem, your help is greatly appreciated.

given $a_n > 0$ and $lim(n a_n) = L$ with $L \ne 0$ show the series $\Sigma a_n$ diverges.

Plato · #2 $Old$ July 3rd, 2009, 04:58 PM

Quote:

Originally Posted by wellfed $View Post$

given $a_n > 0$ and $lim(n a_n) = L$ with $L \ne 0$ show the series $\Sigma a_n$ diverges.

It should be clear to you that $L>0$ .
Use the limit comparsion test and compare $a_n$ with $\frac{1}{n}$ .

Jose27 · #3 $Old$ July 3rd, 2009, 05:01 PM

Quote:

Originally Posted by wellfed $View Post$

Hi I am stuck on the following problem, your help is greatly appreciated.

given $a_n > 0$ and $lim(n a_n) = L$ with $L \ne 0$ show the series $\Sigma a_n$ diverges.

Suppose $0 \leq a_1 \leq ...\leq a_n \leq ...$ and that $\lim_{n \rightarrow \infty} na_n$ $=L$ then $\sum_{i=1}^n {a_i}$ $\leq na_n \leq L$ for all $n \in \mathbb{N}$ and so the series converges.

but then a_n=0 for all n \in \mathbb{N} so it doesn't work. I'll check it later.

halbard · #4 $Old$ July 3rd, 2009, 05:04 PM

Obviously $L>0$ . There is a integer $N$ such that $na_n>L/2$ for all $n\geq N$ . Thus $a_n>L/(2n)$ for all $n\geq N$ , so $\sum a_n$ diverges by comparison with $\sum L/(2n)$ .

wellfed · #5 $Old$ July 3rd, 2009, 05:16 PM

Quote:

Originally Posted by Jose27 $View Post$

Suppose $0 \leq a_1 \leq ...\leq a_n \leq ...$ and that $\lim_{n \rightarrow \infty} na_n$ $=L$ then $\sum_{i=1}^n {a_i}$ $\leq na_n \leq L$ for all $n \in \mathbb{N}$ and so the series converges.

but then a_n=0 for all n \in \mathbb{N} so it doesn't work. I'll check it later.

your assumption $0 \leq a_1 \leq ...\leq a_n \leq ...$ and that $\lim_{n \rightarrow \infty} na_n$ $=L$ are not compatible.

putnam120 · #6 $Old$ July 4th, 2009, 12:29 PM

Just thought I should mention that if $\lim n^\alpha a_n=L$ with $L\neq{0}$ .
Then $\sum a_n<\infty$ if $\alpha>1$ and $\sum a_n$ diverges if $0<\alpha\le{1}$ . Both of these shouldn't be too hard to prove. You basically use the comparison test idea shown in halbard's post.

I haven't looked at the cases where $\alpha<0$ yet.