<s>[INST] Ratio test for $\frac{\sqrt{n^n}}{2^n}$ Before I tried root test, I did ratio test for $\frac{\sqrt{n^n}}{2^n}$ and got: 
$$\lim_{n\to\infty}\frac{\sqrt{(n+1)^{(n+1)}}}{2^{n+1}}\cdot \frac{2^n}{\sqrt{n^n}} = \lim_{n\to\infty} \frac{1}{2}\cdot \sqrt{\frac{(n+1)^{(n+1)}}{n^n}} = \frac{1}{2}$$
But correct answer with ratio test is $\frac{\sqrt{n}}{2}$. So I must have did something wrong
UPDATE: Ratio Test
$$\sqrt[n]{\frac{\sqrt{n}^n}{2^n}}=\frac{\sqrt{n}}{2}$$
 [/INST] The ratio of consecutive terms is indeed $$\frac{1}{2}\cdot \sqrt{\frac{(n+1)^{(n+1)}}{n^n}}$$ but that second factor does not tend to $1$ like you seem to have assumed. We actually have $$ \sqrt{\frac{(n+1)^{(n+1)}}{n^n}} = \sqrt{n+1} \cdot \sqrt{ \left( 1+ \frac{1}{n} \right)^n } .$$
Since $\left( 1+ \frac{1}{n} \right)^n \to e , $  we have that $$ \frac{a_{n+1} }{a_n} \sim  \frac{\sqrt{ne} }{2} $$
which tells you the sequence diverges to infinity. In fact it also tells you $$ a_n\approx \frac{\sqrt{n! e^n} }{2^n} $$
which is also seen more directly if we use $ n! \approx (n/e)^n .$
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