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Proving a series converges

I am doing some self studying in real analysis. I am trying to prove that:
$$\sum _{n=2}^{\infty }\frac {\sqrt {2}^{2n}}{n}$$ converges.
I have tried to use ratio test. However, I get a product of ratios which results in an expression of the form:
$$\frac{\frac{2^{2n+1}}{(n+2)(n-1)}}{\frac{2^{2n+3}n^2}{(n+2)(n-1)(n+1)(n-1)}}$$
But I don’t see how this can be further simplified. Am I missing something? Any help would be appreciated.

A:

The ratio test just asks you to choose two appropriate lower and upper summation bounds for the sum. In this case they can be $n$ and $n+2$. The denominator clearly converges. The numerator can be simplified quite easily:
\begin{align}
\frac{2^{2(n+2)}(n+2)(n-1)}{2^{2n+3}n^2}&=\frac{(n+2)n(n-1)}{n^2}\\
&=\frac{n^2+2n^2-n^2+2n-n^2+4n-2}{n^2}\\
&=\frac{3n}{n^2+4n-2}\\
&
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