Convergence of $\sum_{n=1}^\infty \frac {n^3[\sqrt 2 + (-1)^n]^n}{1.05^n} $ Test convergence of $\sum_{n=1}^\infty \dfrac {n^3[\sqrt 2 + (-1)^n]^n}{1.05^n} $
Attempt: By using the the $n^{th}$ root test : 
$\lim_{n \rightarrow \infty} a_n^{1/n} = \lim_{n \rightarrow \infty} \dfrac {n^{\frac {3}{n}}[\sqrt 2 + (-1)^n] }{1.05 } $

Since, $\lim_{n \rightarrow \infty} n^{\frac {3}{n}}=1$, the above limit has two possible results : $\dfrac {\sqrt 2 +1}{1.05},\dfrac {\sqrt 2 -1}{1.05}$ out of which $\dfrac {\sqrt 2 +1}{1.05}>1$ but $\dfrac {\sqrt 2 -1}{1.05}<1$

In such a case, what can we say about the convergence of $\sum_{n=1}^\infty \dfrac {n^3[\sqrt 2 + (-1)^n]^n}{1.05^n} $?.

If in the denominator, there was $4^n$ instead of $1.05^n$, the series should have been convergent?

Could there be a separate result when there are two limit points of the $n^{th}$ root test?
Thank you very much for your help in this regard.

Denote $\sum a_n$ the given series. We see easily that 
$$\lim_{n\to\infty}a_{2n}\ne0\implies a_n\not\xrightarrow{n\to\infty}0$$
and then the series is divergent.
