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 $$\forall n | ( 1 + 1/(2n) )^n - \sqrt{e} | < |1/n|$$ is true.

$$| ( 1 + 1/(2n) )^n - \sqrt{e} | < | ( 1 + 1/(2n) )^n - e | < \epsilon$$

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 $$\forall n | ( 1 + 1/(2n) )^n - \sqrt{e} | < |1/n|$$ 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 $$|(1 + 1/(2n))^n - \sqrt{e}| < |1/n|$$ 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 $$|1/n| = 1/n < \epsilon$$. That such n being any integer $$> 1/\epsilon$$. Therefore if $$n > 1/\epsilon$$, then $$|(1 + 1/(2n))^n - \sqrt{e}| < \epsilon$$. 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 $$|(1+1/n)^n - e| < \epsilon$$. 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 $$\lim_{n\rightarrow +\infty} {(1+1/(2m))^m } = \sqrt{e}$$.