Sunday, 31 May 2015

real analysis - Converge Test on the series sumlimitsinftyn=0left(frac2n+n334nright)n



I want to show, that a:=n=0(2n+n334n)n is not converging, because lim. Therefore, the series can't be absolute converge too.



Firstly, I try to simplify the term. After that I want to find the limit.




Unfortunately, I can't seem to find any good equation with that I can clearly show (*).
\begin{align} \sum \limits_{n=0}^{\infty} \left(\dfrac{2n+n^3}{3-4n}\right)^n&=\left (\dfrac{\not{n}\cdot (2+n^2)}{\not n \cdot (\frac{3}{n}-4)}\right)^n\\ &=\left(\dfrac{2+n^2}{\frac 3n-4}\right)^n\\ &= \cdots \end{align}



How to go on?


Answer



Recall that




\sum_0^\infty a_n <\infty \implies a_n \to 0



therefore if a_n \not \to 0 the series can’t converge.



In that case for n\ge 3 we have



\left|\dfrac{2n+n^3}{3-4n}\right|=\dfrac{2n+n^3}{4n-3}>\dfrac{n^3}{4n}=\frac{n^2}4\ge2



and then




|a_n|=\left|\dfrac{2n+n^3}{3-4n}\right|^n\ge 2^n



Refer also to the related:




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}...