Does the series $\sum \limits _{n=1}^\infty \frac{1}{n} \left[{\frac{1}{(-1)^n-5}}\right]^{n}$ converge or diverge? $$\sum \limits _{n=1}^\infty \frac{1}{n} \left[{\frac{1}{(-1)^n-5}}\right]^{n}$$
I applied the root test and believe that this series converges. That is, I found that:
$$\text{limsup} \left|{\frac{1}{n} \left({\frac{1}{(-1)^n-5}}\right)^n}\right|^{\frac{1}{n}} <1$$
Can someone verify that I am correct. No need for a long proof. Thanks!
 A: The series $\sum \limits _{n=1}^\infty \frac{1}{n} \left({\frac{1}{(-1)^n-5}}\right)^n$ is absolutely convergent since $$\frac{1}{n}\cdot \left|{\frac{1}{(-1)^n-5}}\right|^n \leqslant \frac{1}{n}\cdot\dfrac{1}{4^n}.$$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\sum_{n=1}^{\infty}{1 \over n}\bracks{1 \over \pars{-1}^{n} - 5}^{n}
=
\sum_{n = 0}^{\infty}\braces{%
{1 \over 2n + 2}\,\bracks{1 \over \pars{-1}^{2n + 2} - 5}^{2n + 2}
+
{1 \over 2n + 1}\,\bracks{1 \over \pars{-1}^{2n + 1} - 5}^{2n + 1}}
\\[3mm]&=
\sum_{n = 0}^{\infty}\pars{%
{4^{-2n - 2} \over 2n + 2} - {6^{-2n - 1} \over 2n + 1}}
\end{align}

$\ds{\fermi\pars{x} \equiv \sum_{n = 0}^{\infty}{x^{2n + 2} \over 2n + 2}\,,
     \quad
     {\rm g}\pars{x} \equiv \sum_{n = 0}^{\infty}{x^{2n + 1} \over 2n + 1}\,,
     \quad\verts{x} < 1\,,\quad\fermi\pars{0} = {\rm g}\pars{0} = 0}$.
\begin{align}
\fermi'\pars{x} &= \sum_{n = 0}^{\infty}x^{2n + 1} = {x \over 1 - x^{2}}
=-\,{1 \over 2}\,\pars{{1 \over x - 1} + {1 \over x + 1}}
\\[3mm]\imp &\fermi\pars{x}
= -\,{1 \over 2}\,\ln\pars{\bracks{x - 1}/\bracks{x + 1} \over -1}
= -\,{1 \over 2}\ln\pars{1 - x \over 1 + x}
= {1 \over 2}\ln\pars{1 + x \over 1 - x}
\\&\mbox{Similarly,}
\\
{\rm g}\pars{x} &= \sum_{n = 0}^{\infty}x^{2n} = {1 \over 1 - x^{2}}
={1 \over 2}\,\pars{{1 \over x + 1} - {1 \over x - 1}}
\\[3mm]\imp &{\rm g}\pars{x}
= {1 \over 2}\,\ln\pars{\bracks{x + 1}/\bracks{x - 1} \over -1}
= {1 \over 2}\ln\pars{1 + x \over 1 - x} = \fermi\pars{x}
\end{align}

$$
\sum_{n=1}^{\infty}{1 \over n}\bracks{1 \over \pars{-1}^{n} - 5}^{n}
=
\fermi\pars{1 \over 4} - \fermi\pars{1 \over 6}
=
{1 \over 2}\,\ln\pars{5 \over 3} - {1 \over 2}\,\ln\pars{7 \over 5} 
$$
$$\color{#0000ff}{\large%
\sum_{n=1}^{\infty}{1 \over n}\bracks{1 \over \pars{-1}^{n} - 5}^{n}
\color{#000000}{\ =\ }
{1 \over 2}\,\ln\pars{25 \over 21}} \approx 0.0872
$$
A: Compare your series to the alternating harmonic series to see that it converges
