Prove non-existence of $\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}$

Question

I was trying for a while to prove non-existence of theu following limit:

$$\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}$$

Unfortunately, with no results.

My hope was to show, that:

$$\lim_{n\to\infty}\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}\not=0$$

But showing that was harder than I thought.

Can anyone show me how to solve this problem?

Answer 1

let $\sqrt[n]{5}=x$ $$\lim_{n\to\infty}\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}=\lim_{n\to\infty}\frac{2^n+4n+6}{2^n(x-1)}=\lim_{n\to\infty}\frac{(2^n+4n+6)\cdot(1+x+x^2+\cdots+x^{n-1})}{2^n(x^n-1)}$$ now substitute $x$ and get your answer to be non-zero

Answer 2

Note first that $$ \lim_{n\to\infty}n(\sqrt[n]{5}-1)=\ln 5, $$ Hence $$ (-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}=n\cdot(-1)^n\cdot\frac{1+\frac{4n}{2^n}+\frac{6}{2^n}}{n(\sqrt[n]{5}-1)}. $$ Clearly $$ \frac{1+\frac{4n}{2^n}+\frac{6}{2^n}}{n(\sqrt[n]{5}-1)}\to \frac{1}{\ln 5} $$ while $$ n\cdot(-1)^n\,\,\,\text{diverges}. $$ Hence, so does $$ n\cdot(-1)^n\cdot\frac{1+\frac{4n}{2^n}+\frac{6}{2^n}}{n(\sqrt[n]{5}-1)}. $$

Answer 3

Try to show difference bwtween $\lim_{n \rightarrow \infty} \sup a_{n}$ and $\lim_{n \rightarrow \infty} \inf a_{n}$.

Answer 4

$$\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}=\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n}\frac1{\sqrt[n]{5}-1}$$

is of the form $\dfrac10$.

Answer 5

Hint:Divide the numerator and denominator by $2^n$. You dont need Lhopital rule .