Need guidance on my proof of limit.

[Samuelb88]

Homework Statement

Show that lim_{n\rightarrow +\infty} { (1 + 1/(2n))^n } = \sqrt{e}.

Homework Equations

I am allowed to assume * lim_{n\rightarrow +\infty} { (1 + 1/n )^n} = e.
I am not allowed to use the theorem that asserts lim_{n\rightarrow n_0} {\sqrt{S_n}} = (lim_{n\rightarrow n_0} {S_n})^{1/2}.

The Attempt at a Solution

I want to show that the sequence is increasing and bounded, and therefore the lim_{n\rightarrow +\infty} { (1 + 1/(2n))^n } exists. Let's suppose I have show the sequence is increasing by comparing S_k and S_{k+1} and showing \forall k, S_k < S_{k+1}. Let's also suppose I know that 2 is an upper bound for S_n. Then I want to show \forall \epsilon >0 \exists N such that \forall n > N, | (1 + 1/(2n) )^n - \sqrt{e} | < \epsilon.

Does this argument work?

Lemma: \forall \epsilon > 0 \exists N_0 such that \forall n > N_0, |2 + 1/n - 2| < \epsilon. That is, lim_{n\rightarrow +\infty} {(2 + 1/n)} = 2.

\forall n | ( 1 + 1/(2n) )^n - \sqrt{e} | < |1/n|. Since \forall n, 1/n > 0, then |1/n| = 1/n. By our lemma, we know |1/n| = 1/n < \epsilon. Choose an n > 1/ \epsilon. Then | ( 1 + 1/(2n) )^n - \sqrt{e} | < \epsilon and therefore lim_{n\rightarrow +\infty} { (1 + 1/(2n))^n } = \sqrt{e}.

I was originally trying to use * with the definition of a limit to show | ( 1 + 1/(2n) )^n - \sqrt{e} | < | ( 1 + 1/(2n) )^n - e | < \epsilon but I couldn't figure out how to determine N.

 

[Office_Shredder]

It's not clear to me what the relevance of the lemma is to picking your value of n (why can't you just pick n larger than one over epsilon without looking at a different limit?), or why <br /> \forall n | ( 1 + 1/(2n) )^n - \sqrt{e} | &lt; |1/n|<br /> is true.

<br /> | ( 1 + 1/(2n) )^n - \sqrt{e} | &lt; | ( 1 + 1/(2n) )^n - e | &lt; \epsilon <br />

This middle term in the inequality is going to fairly large (in the analysis sense) when n is large. We know ( 1 + 1/(2n) )^n is going to be close to \sqrt{e}, so it can't be close to e as well. However, at some point you are going to need to use the limit that gives e, or some alternative definition of e

 

[Samuelb88]

It's not clear to me what the relevance of the lemma is to picking your value of n (why can't you just pick n larger than one over epsilon without looking at a different limit?), or why \forall n | ( 1 + 1/(2n) )^n - \sqrt{e} | &lt; |1/n| is true.

I added the lemma to use the fact I knew the limit of the 2+1/n was 2, and to bound the difference |(1 + 1/(2n))^n - \sqrt{e}| &lt; |1/n| for all n (n is of course restricted to naturals, and this doesn't include zero). So my reasoning was after establishing the above inequality, to use the lemma to argue I can find an n>N such that |1/n| = 1/n &lt; \epsilon. That such n being any integer &gt; 1/\epsilon. Therefore if n &gt; 1/\epsilon, then |(1 + 1/(2n))^n - \sqrt{e}| &lt; \epsilon. Does this make sense?

This middle term in the inequality is going to fairly large (in the analysis sense) when n is large. We know ( 1 + 1/(2n) )^n is going to be close to \sqrt{e} , so it can't be close to e as well. However, at some point you are going to need to use the limit that gives e, or some alternative definition of e

Sorry, I meant to say: |(1 + 1/(2n)^n - \sqrt{e}| < |(1+1/n)^n - e|. Then using the fact I that I knew \forall \epsilon there was some N such that whenever n>N, I was trying to argue |(1+1/n)^n - e| &lt; \epsilon. The main problem I had was deriving an explicit N from the inequality above. That is why I choose to bound the difference |(1 + 1/(2n)^n - \sqrt{e}| by 1/n instead.

 

[Samuelb88]

Here is an idea that I just thought of using the fact that I may assume * \lim_{n/rightarrow +/infty} {(1+1/n)^n}=e. I am not sure if it is considered "rigorous" enough. Here it is:

I may use the fact that ** if S_n, T_n are sequences, and \lim_{n\rightarrow +\infty} {S_n} =A, \lim_{n\rightarrow} {T_n} = B. Then \lim_{n\rightarrow} {S_n T_n} = AB.

Suppose * and let m=2n. Then \lim_{n\rightarrow +\infty} {(1 + 1/n)^n} = \lim_{m\rightarrow +\infty} {(1+1/(2m))^{2m} (Here I am making the argument that as n\rightarrow +\infty, 2m\rightarrow +\infty and hence m\rightarrow +\infty). By **, \lim_{n\rightarrow +\infty} {(1+1/(2m)^m (1+1/(2m))^m } = \lim_{m\rightarrow +\infty} {(1+1/(2m))^m} \lim_{m\rightarrow +\infty} {(1+1/(2m))^m = e. Then \lim_{n\rightarrow +\infty} {(1+1/(2m))^m } = \sqrt{e}.