I have the following question.. \begin{align*} f(n) &= n^{\sqrt{n}}\\ g(n) &= 2^{an},\quad \text{where $a > 1$} \end{align*} Evaluate the limit of $f(n)/g(n)$ as $n \to \infty$
I can intuitively see this has to be $0$, since $2^{an}$ grows much faster than $n^{\sqrt{n}}$, but how to show this rigorously?
You may also apply the root test for infinite series and get
$$\displaystyle\sqrt[n]{\dfrac{f(n)}{g(n)}} = 2^{-a}n^\dfrac{1}{\sqrt{n}}\stackrel{n\to\infty}{\longrightarrow}2^{-a}\stackrel{a>1}{<}1\tag1$$
Since $a>1$ you have $2^{-a}<1$
Thus the series $\displaystyle\sum_{n\in\mathbb N} \dfrac{f(n)}{g(n)}$ converges, which implies, that
$$\dfrac{f(n)}{g(n)}\stackrel{n\to\infty}\longrightarrow 0 $$
To verify the limit in (1) one can apply L'Hospital ($\star$) rule as follows
$$\lim_{n\to\infty}n^\dfrac{1}{\sqrt{n}} = \exp{\log \lim_{n\to\infty}n^\dfrac{1}{\sqrt{n}}}=\exp{\lim_{n\to\infty}\dfrac{\log n}{\sqrt{n}}}\stackrel{\star}{=} \exp \lim_{n\to\infty}\dfrac{2}{\sqrt{n}}=\exp0=1$$
You should be able to use the ratio test for sequences. That is, evaluate $$\lim_{n \to \infty} \frac{(n+1)^{\sqrt{n+1}}}{2^{a(n+1)}}*\frac{2^{an}}{n^{\sqrt{n}}}$$
and you should get $\frac{1}{2^a} <1$ (why?) implying that the sequence has a zero limit as $n$ tends to infinity.
