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:

1. 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.

2. 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 $\lim{(a_n)}\geq e$.

3. $\lim{(a_n)}\geq e^2$ 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.]

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After the first line of step #1, you have an explicit expression for $a_n$. Just use Stirling's formula to find out its limit… – F'x Jan 28 '13 at 20:18

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 $n\to\infty$.

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Hint: Let $n=100$. Then half the terms are $\ge 1.5$.

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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 $\lim a_n= \infty$.

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Sorry, misread the question. $a_n>(1.5)^\frac{n}{2}$ So it diverges. You could also use Stirling on the factorial.

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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.

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@user1709828: we like to work together :-) – user 1618033 Jan 28 '13 at 14:26

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.

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I am wrong. I will fix it using $\int_1^2 \log t\,dt = 2\log 2 - 1$. – GEdgar Apr 20 '13 at 20:11
Thanks a lot sir.Got it. – learner Apr 21 '13 at 4:32

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.

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