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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


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The update looks like the root test, not the ratio test. – robjohn Mar 11 '12 at 10:49
up vote 4 down vote accepted

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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The answer cannot be $\sqrt{n}/2$ (it depends on $n$), but $\infty$: $$ \frac{(n+1)^{(n+1)}}{n^n}=(n+1)\Bigl(1+\frac1n\Bigr)^n $$

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