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I am trying to prove that the sequence $a_n=(1+\frac{1}{n})^n$ converges by proving that it is a Cauchy sequence.

I don't get very far, see: for $\epsilon>0$ there must exist $N$ such that $|a_m-a_n|<\epsilon$, for $ m,n>N$

$$|a_m-a_n|=\bigg|\bigg(1+\frac{1}{m}\bigg)^m-\bigg(1+\frac{1}{n}\bigg)^n\bigg|\leq \bigg|\bigg(1+\frac{1}{m}\bigg)^m\bigg|+\bigg|\bigg(1+\frac{1}{n}\bigg)^n\bigg|\leq\bigg(1+\frac{1}{m}\bigg)^m+\bigg(1+\frac{1}{n}\bigg)^n\leq \quad?$$

I know I am supposed to keep going, but I just can't figure out the next step. Can anyone offer me a hint please? Or if there is another question that has been answered (I couldn't find any) I would gladly look at it.

Thanks so much!

user66807
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  • http://math.stackexchange.com/questions/64860/proving-bigl1-frac1n1-bigrn1-gt-1-frac1nn/ –  Mar 19 '13 at 02:10
  • @ByronSchmuland I looked at the link, but can you explain what the AM-GM inequality is, and what $\prod_{i=0}^n , x_i .$ mean? Thanks! – user66807 Mar 19 '13 at 02:30
  • @ Sebastian Griotberg (I don't know if it matters or not, but this isn't a homework question. Its for preparation for a test) – user66807 Mar 19 '13 at 03:27

2 Answers2

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We have the following inequalities:

$\left(1+\dfrac{1}{n}\right)^n = 1 + 1 + \dfrac{1}{2!}\left(1-\dfrac{1}{n}\right)+\dfrac{1}{3!}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \ldots \leq 2 + \dfrac{1}{2} + \dfrac{1}{2^2} + \dots =3$

Similarly,

$ \begin{align*} \left(1-\dfrac{1}{n^2}\right)^n &= 1 - {n \choose 1}\frac{1}{n^2} + {n \choose 2}\frac{1}{n^4} + \dots\\ &= 1 - \frac{1}{n} + \dfrac{1}{2!n^2}\left(1-\dfrac{1}{n}\right) - \dfrac{1}{3!n^3}\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) + \ldots \end{align*} $

So, $$ | \left(1-\dfrac{1}{n^2}\right)^{n} -\left(1-\frac{1}{n} \right)| \leq \dfrac{1}{2n^2} + \dfrac{1}{2^2n^2} + \dfrac{1}{2^3n^2} + \ldots = \dfrac{1}{n^2} . $$

Now,

$ \begin{align*} \left(1+\frac{1}{n+1}\right)^{n+1} - \left(1+\frac{1}{n}\right)^n &= \left(1+\frac{1}{n}-\frac{1}{n(n+1)}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n \\ &=\left(1+\frac{1}{n}\right)^{n+1}\left\{ \left( 1- \frac{1}{(n+1)^2}\right)^{n+1} - \frac{n}{n+1}\right\}\\ &= \left(1+\frac{1}{n}\right)^{n+1}(1 - \frac{1}{n+1} + \text{O}(\frac{1}{n^2}) - \frac{n}{n+1})\\ & = \left(1+\frac{1}{n}\right)^{n+1}\text{O}(\frac{1}{n^2}) \\ &= \text{O}(\frac{1}{n^2}) \text{ (since $(1+1/n)^{n+1}$ is bounded) }. \end{align*} $

So, letting $a_k = (1+1/k)^k$ we have, $|a_{k+1}-a_k| \leq C/k^2$ for some $C$ and hence, $\sum_{ k \geq n } | a_{k+1} - a_k | \to 0$ as $n \to \infty$.

Since $|a_n - a_m| \leq \sum_{ k \geq \min\{m,n\}} |a_{k+1} - a_k|$. So given $\epsilon > 0$ chose $N$ such that $\sum_{ k \geq N} |a_k - a_{k+1}| < \epsilon$ and $|a_n - a_m| < \epsilon $ for $n,m \geq N$.

Arin Chaudhuri
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Look if it works!
For $m>n \geq 1, |a_m-a_n| \leq |a_m^{1/m}-a_n^{1/n}|=|\frac {1}{m}-\frac {1}{n}|<\frac {1}{m}+\frac {1}{n}<\frac {2}{n}< \epsilon$ if $n >[2 /\epsilon].$

learner
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  • That idea came to me, but I didn't think of using it at the beginning. Thanks so much, very helpful. – user66807 Mar 19 '13 at 03:47
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    @learner, how do you prove the very first inequality from the left? – DonAntonio Mar 19 '13 at 03:51
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    @DonAntonio That's a good question that I hadn't thought of. I don't know how to prove it, but plugging in values where $m>n\geq1$ works, but proving it would be helpful as well. – user66807 Mar 19 '13 at 04:12
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    Try plugging in $n=5$ and $m=7$. –  Mar 19 '13 at 15:34
  • @ByronSchmuland You are right. When plugging in those values the inequality doesn't hold...unless of course I didn't calculate properly. – user66807 Mar 29 '13 at 21:29