Show $a_n$ is unbounded if $a_n= a_{n-1} \left(1+ \frac{1}{\sqrt n}\right)$ to determine that $a_n$ diverges. I wish to determine the limit of $a_n$, where we recursively define:

$$a_n= a_{n-1} \left(1+ \frac{1}{\sqrt n}\right)$$ where $a_0=1$

I already noticed it is increasing, because 
$$a_n - a_{n-1}= \frac{1}{\sqrt{n}}a_{n-1}>0 $$
Since all terms are strictly positive. (alternatively, we could form a better argument via induction). 

edit:
Base case: $a_1 = 2 > 1 = a_0$. 
Hypothesis: $a_k > a_{k-1}$
Step: $a _ {k+1}= a_{k} \left(1+ \frac{1}{\sqrt n}\right)>a_{k-1} \left(1+ \frac{1}{\sqrt n}\right)=a_k $
Therefore $a_n >a_{n-1}$ for all $n$ and the sequence is increasing.

I plotted some terms and this seems to grow exponentially, how would I prove it diverges? I think I need to show it is unbounded.I tried the ratio test for sequences but this is inconclusive as the ratio tends to $1$.
 A: Further to my comment, starting with $$a_n = a_{n-1}\left(1+\frac{1}{\sqrt{n}}\right)=a_{n-2}\left(1+\frac{1}{\sqrt{n-1}}\right)\left(1+\frac{1}{\sqrt{n}}\right)=\\
a_0\prod\limits_{k=1}^n\left(1+\frac{1}{\sqrt{k}}\right)\geq \left(1+\frac{1}{\sqrt{n}}\right)^n=
\left(\left(1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}} \geq ...$$
and applying Bernoulli's inequality
$$...\geq \left(1+\frac{\sqrt{n}}{\sqrt{n}}\right)^{\sqrt{n}}=2^{\sqrt{n}}$$
As a result
$$a_n \geq 2^{\sqrt{n}}$$
and the result follows, i.e. the sequence is unbounded and diverges.
A: Use $\sqrt n + 1 \geq \sqrt{n+1}, n\in\mathbb N:$
$$a_n = a_{n-1}(1+\frac 1{\sqrt n})\implies a_n = \prod_{k=1}^n(1+\frac 1{\sqrt k})= \prod_{k=1}^n \frac{\sqrt k + 1}{\sqrt k} \geq \prod_{k=1}^n\frac{\sqrt{k+1}}{\sqrt k} = \sqrt{n+1}.$$
A: $a_n=\prod_{i=1}^n\left(1+\frac1{\sqrt{i}}\right).$  Now notice that $f(x)=1+x-e^{x/4}\ge 0$ for $x\leq 1.$  Hence $a_n\geq e^{\frac14\sum_{i=1}^n\frac1{\sqrt{i}}}\rightarrow \infty.$
A: Another approach:
$$a_n=a_{n-1}+\frac{1}{\sqrt{n}}a_{n-1}\geq a_{n-1}+\frac{1}{\sqrt{n}},$$
from which we quickly get
$$a_n\geq\sum_{k=1}^n\frac{1}{\sqrt{k}}\geq\sum_{k=1}^n\frac{1}{k}$$
which is divergent, since the last sum gives the harmonic series.
