I want to show, that a:=∞∑n=0(2n+n33−4n)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:
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