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.

For a series to converge, any sub-series must converge as well.
Since you have shown that there exists a divergent sub-series by the root test, it follows that the series diverges. 
As to your second question, the answer is no. Since you take the $n^{th}$ root, the limit always goes to zero.
