$\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!
 A: 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.
A: (Expanding my comment into an answer)
Here's a broad hint: $\left(\dfrac{1-2n}{6+2n}\right) = -\left(\dfrac{2n-1}{2n+6}\right) = -\left(1-\dfrac{7}{2n+6}\right)$, so your general term is $(-1)^n\left(1-\dfrac{7}{2n+6}\right)^n$.  Now, set $y=2n+6$ (so $n=\frac12(y-6)=\frac12y-3$); then the positive piece of this term (setting aside the $(-1)^n$ for the moment) is $\left(1-\frac{7}{y}\right)^{\frac12y-3} = \left(1-\frac7y\right)^{-3}\cdot\left(\left(1-\frac7y\right)^y\right)^{\frac12}$.  The inner portion of the right hand side of this expression should look a little familiar...
A: Notice that $\displaystyle\lim_{n\to\infty}\bigg(\frac{2n-1}{2n+6}\bigg)^n=\lim_{n\to\infty}\bigg(\frac{1+\frac{-\frac{1}{2}}{n}}{1+\frac{3}{n}}\bigg)^n=\lim_{n\to\infty}\frac{(1+\frac{-\frac{1}{2}}{n})^n}{(1+\frac{3}{n})^n}=\frac{e^{-\frac{1}{2}}}{e^3}=e^{-\frac{7}{2}}\ne0$,
so the series diverges.
