$\sum (\frac{1-2n}{6+2n})^n $ converges? Verify if
$$\sum_{n=0}^{\infty} \left(\frac{1-2n}{6+2n}\right)^n $$
converges
The root test is inconclusive and the limit of the general term is 0. I think I should use the comparison test, in this case. But I couldnt find any functiob to use. Any hint?
Thanks!

Let $a_n=\left(\dfrac{1-2n}{6+2n}\right)^n$. We have that
$$|a_n|=\left|\dfrac{2n-1}{6+2n}\right|^n=\exp\left[n \log \left(\dfrac{2n-1}{6+2n}\right)\right].$$
Let $b_n=n \log \left(\dfrac{2n-1}{6+2n}\right)$. Making the substitution $n=\dfrac{1}{x}$, we have
$$\lim_{n\to\infty}b_n=\lim_{x\to0}\dfrac{1}{x}\log\left(\dfrac{2-x}{2+6x}\right)=\lim_{x\to0}\dfrac{1}{x}\log\left(1-\dfrac{7x}{2+6x}\right)=\lim_{x\to0}\dfrac{1}{x}\left(-\dfrac{7x}{2+6x}\right)=-\dfrac{7}{2}.$$
Hence,
$$\lim_{n\to \infty}|a_n|=\lim_{n\to \infty} e^{b_n}=e^{-\tfrac{7}{2}},$$
and the series diverges by the divergence theorem.
