Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.]

share|improve this question
    
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

7 Answers 7

up vote 8 down vote accepted

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 $n\to\infty$.

share|improve this answer

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

share|improve this answer

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 $\lim a_n= \infty$.

share|improve this answer

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

share|improve this answer

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.

share|improve this answer
1  
@user1709828: we like to work together :-) –  Chris's sis 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.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.