Friday, 4 October 2013

real analysis - Counterexample: If suminftyn=1an=1 and each angeq0, then limlimitsntonan=0



I'm studying for an exam and I ran into these problems. I'm having a feeling that this is not true. Hence, I don't need to prove. I need to just provide a counterexample. However, the appropriate example is not just coming. Any help?



Prove or give a counterexample:




  1. If n=1an=1 and each an0, then lim



  2. If a_n\geq 0 and \sum_{n=1}^{\infty}a_n converges, then \lim\limits_{n\to\infty}na_n=0



Answer



Let a_{n^2}=\dfrac{1}{n^2} and a_n=0 otherwise. Then na_n does not tend to 0, and the series is convergent.



The sum is \dfrac{\pi^2}{6}, but note you can multiply each term by the same constant to make the limit any nonzero number you wish.






Note, however, that if a_n is also nonincreasing, the situation is different: Series converges implies \lim{n a_n} = 0



No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...