Convergence of the sequence $(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})$ 
I have a sequence $(a_n)$ where for each natural number $n$,
  $$a_n = (1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})$$ and I want to find its limit as $n\to\infty$.

I obviously couldn't prove it and after several futile attempts decided to post it here. 
Here is a list of a few observations which I got from those attempts:


*

*The sequence $(a_n)$ is a strictly increasing sequence. To prove this, I rewrote each element as $$a_n = (1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})= \frac{(n+1)\cdots(n+n)}{n^n}= \frac{(2n)!}{n!n^n}.$$
Then $$\frac{a_{n+1}}{a_n}= \frac{2(n+1)!}{(n+1)!(n+1)^{n+1}}\frac{n!n^n}{(2n)!}=\frac{(2n+1)(2n+2)}{(n+1)^2(1+\frac{1}{n})^n} \to \frac{4}{e}$$ as $n\to\infty$. Since $\frac{4}{e}>1$ we have $a_{n+1}>a_n$ eventually.

*The limit of this sequence is bounded below by $e$. By replacing $1,2, \ldots, n$ with $1$ in the expression of $a_n$, we get $a_n \geq (1+\frac{1}{n})^n$. And thus $\lim{(a_n)}\geq e$. 

*$\lim{(a_n)}\geq e^2$ and $\lim{(a_n)}\geq e^3$. The first assertion follows from the fact that $$a_n\geq(1+\frac{1}{n})(1+\frac{2}{n})^{n-1}= \frac{(1+\frac{1}{n})(1+\frac{2}{n})^{n}}{(1+\frac{2}{n})} \to e^2.$$ And the last one follows the same way because $$a_n\geq (1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})^{n-2}.$$
Now I have a gut feeling that for any natural number $k$, one can show that for all sufficiently large natural number $n$, $$a_n\geq (1+\frac{1}{n})\cdots(1+\frac{k-1}{n})(1+\frac{k}{n})^{n-(k-1)}.$$ And therefore for all $k \in \mathbb{N}$, $\lim{(a_n)}\geq e^k$ making the sequence divergent. But I'm really not sure about this approach and I'll appreciate any help towards this end. Thank you.
[Note: As this sequence is quite common, there may be other posts on math.SE asking the same question. I didn't search for them because I just don't know how to search for an expression this big. Though a link related to any previous question concerning this particular sequence will be good enough, I will greatly appreciate if someone takes the trouble to look into my approach/observations and point out where  I'm going wrong.]
 A: Hint: Let $n=100$. Then half the terms are $\ge 1.5$.
A: Sorry, misread the question. $a_n>(1.5)^\frac{n}{2}$ So it diverges. You could also use Stirling on the factorial.
A: Good job, why didn't you do a last step? As you observed, $a_n$ is monotonic and  $\lim a_n \geq \mathrm e^k$ for any $k$ which is more that enough to conclude that $\lim a_n= \infty$.
A: Let $a_n$ be a sequence of positive numbers. Then, if $\prod_{n=1}^{\infty} (1+a_n)$ converges/diverges, then
$\sum_{n=1}^{\infty} a_n$ converges/diverges (and the converse is also true).
Sister.
A: Another divergence proof.  In fact
$$
\frac{\log a_n}{n} = \frac{1}{n}\sum_{k=1}^n \log\left(1+\frac{k}{n}\right)
$$
converges to $\int_1^2\log t\,dt = 2\log 2 - 1$, a Riemann sum argument.  And therefore $\log a_n$ goes to infinity and $a_n$ goes to infinity.
A: If you multiply it out and throw out most of the terms, you end up with
$$a_n > \frac1n+\frac2n+\frac3n+\cdots+\frac nn=\frac1n\sum_{k=1}^n k=\frac{n+1}2\to\infty$$
as $n\to\infty$.
A: The sequence $\prod_{n \ge 0} (1 + a_n)$ converges if and only if $\sum_{n \ge 0} a_n$ converges. See for example http://cornellmath.wordpress.com/2008/01/26/convergence-of-infinite-products. In ths case the sum is the harmonic series, and that one diverges.
