Is $n^{1/\sqrt{n}}$ convergent? I think it is convergent to $1$ because as $n$ tends to $\infty$ , $1/\sqrt(n)$ tends to $0$. Is it true?
Thanks!
 A: $$n^{\frac1{\sqrt n}}\stackrel{\text{def}}{=}\mathrm e^{\frac{\log n}{\sqrt n}}.$$
Now a basic limit is $\;\lim_{n\to\infty}\dfrac{\log n}{n}=0$, from which we deduce, for any $\alpha>0$: $$\;\lim_{n\to\infty}\dfrac{\log(n^\alpha)}{n^\alpha}=\alpha\lim_{n\to\infty}\dfrac{\log n}{n^\alpha}=0\;$$ 
whence $\;\lim_{n\to\infty}\dfrac{\log n}{n^\alpha}=0$. Thus, the exponent of $\mathrm e$ tends to $0$, and $n^{\frac1{\sqrt n}}$ tends to $1$.
A: Let $y_n=n^{\frac{1}{\sqrt n}}$ so we have  $\log y_n=\frac{\log n}{\sqrt n}$.
Applying now the Hospital rule you have $$\lim_{n\to \infty}y_n=\lim_{n\to \infty}\frac{\log n}{\sqrt n}=\lim_{n\to \infty}\frac{\frac 1n}{-2\sqrt{n^3}}=0$$
Thus your limit is equal to $1$ so your sequence is convergent to $1$ as you think.
A: $$ n^{\left( \frac{1}{\log n} \right)} = e  $$
$$ \lim_{n \rightarrow \infty} n^{\left( \frac{1}{ \log \log n} \right)} = \infty  $$
A: for $x \in \mathbb{R^+} $let 
$$
A_x=x^{1/\sqrt{x}}
$$
giving
$$
\log A_x = \frac2{\sqrt{x}}\log \sqrt{x}
$$
differentiating $\log A_x$ wrt $x$
$$
\frac{A'_x}{A_x} = x^{-\frac32}(1-\log \sqrt{x})
$$
so $A_x$ is a decreasing function of $x$ for $x \gt e^2$ and is bounded below by 1. This implies that a limit exists as $x \to \infty$, and
$$
\lim_{n \to \infty} \log A_n = \lim_{n^2 \to \infty} \log A_{n^2}
$$
set $N=n^2$,
$$ 
\log A_{N} = \frac{2\log n}{n}
$$
since $\lim_{n^2 \to \infty} \frac{\log n}n =0 $ we have $\log A_n \to 0$ and $A_n \to 1$
