Question

Real Analysis Help!
My teacher wanted us to look over a proof and understand it, but I'm having trouble understanding the second part.

Answer 1

Corollary 1 The limit
\[e= (1+ \frac{ 1 }{ n })^n\]
exists.


Proof

Let
\[e_{n}= (1+ \frac{ 1 }{ n })^n = (\frac{ n+1 }{ n })^n\]
** I understand this step, you just have to get a common donominator.

***(I don't understand what is happening from this point on.)
\[\frac{ e_{n} }{ e_{n}-1 }= (\frac{ n+1 }{ n })^n (\frac{ n }{ n-1 })^n (\frac{ n }{ n-1 })=\]
\[(\frac{ n+1 }{ n }* \frac{ n-1 }{ n })^n (\frac{ n }{ n-1 })=\]
\[(1- \frac{ 1 }{ n^2 })^n (\frac{ n }{ n-1 })>(1-\frac{ n }{ n^2 }) (\frac{ n }{ n-1 })=1\]

where we have used Bernoulli's inequality at the last step. Thus.,
\[e_{n-1} <e_{n}, \]
so that
\[e_{n}\] is increasing, and \[e_{n} >e_{1}= 2\]

Answer 2

is it \(e_{n-1}\) or is it \(e_n-1\) ?

Answer 3

there is a mistake somewhere; it is ((n-1)/n)^n; not the inverse

Answer 4

i looks to me like you are trying to show that \[\lim_{n\to \infty}\frac{e_n}{e_{n-1}}=1\] maybe i am wrong

Answer 5

so that
\[\frac{e_n}{e_{n-1}}=\frac{(\frac{n+1}{n})^n}{(\frac{n}{n-1})^{n-1}}\] and then algebra

Answer 6

im assuming n is a natural number...

Answer 7

rewrite as
\[\frac{e_n}{e_{n-1}}=(\frac{n+1}{n})^n(\frac{n-1}{n})^{n-1}\]

Answer 8

then multiply and divide by \(\frac{n-1}{n}\) to get the exponents equal for the first term

Answer 9

get
\[\frac{e_n}{e_{n-1}}=(\frac{n+1}{n})^n(\frac{n-1}{n})^n\frac{n}{n-1}\]

Answer 10

as for bernoulli, i am only proficient in elementary algebra

Answer 11

I am taking real analysis myself so hopefully when I see this I can do it haha

Answer 12

Sorry I didn't realize it continued onto the next page.

Continuation:

To show that the sequence \[e_{n}\] is bounded, observe that by the Binomial Theorem and Part (c) of Problem 7 Section 1.2,
\[e_{n}= (1+ \frac{ 1 }{ n })^n= \sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right) \frac{ 1 }{ n^k }\le \sum_{k=0}^{n}\frac{ 1 }{ k! }\]
\[=1+1+\frac{ 1 }{ 2! }+\frac{ 1 }{ 3! }+...+\frac{ 1 }{ k! }\]
\[<1+1+\frac{ 1 }{ 2 }+\frac{ 1 }{ 2^2 }+...+\frac{ 1 }{ 2^k }<3\]
Therefore, \[2<e_{n}<3\] is an increasing sequence. It follows that the limit
\[e= \lim (1+ \frac{ 1 }{ n })^n\] exists and is a number between 2 and 3. Actually, of course, e=2.71828...to five places. We will see how to compute e in Chapter 8.

Answer 13

As far as I can see I copied this down correctly from the book.

Answer 14

which real analysis book do you guys use? Bartle?
I am pretty sure the book made a mistake in equation 3. my algebra looks right so I am not the one making the error

Answer 15

Actually no I lied. I caught one mistake:

\[\frac{e _{n} }{ e_{n}-1 }= (\frac{ n+1 }{ n })^n(\frac{ n }{ n-1 })^{-n} (\frac{ n }{ n-1 })\]

Answer 16

We use Basic Real Analysis by James S. Howland (International Series in Mathematics)

Answer 17

oh nice. we just started limits today in my class but I am still confused since we never specified what n approaches haha. well now that you caught the mistake I think you are good right?

Answer 18

This class confuses me to no end. I don't understand half of what we're doing. I think I may just ask my teacher to explain it. I really wished our book was step by step.

Answer 19

I can't see Part (c) of Problem 7 Section 1.2 of your book so I can't explain the continuation part. all I can do is that pdf file I attached about the first part. Real Analysis is really confusing; my teacher doesn't even explain things step by step... good luck :)

Answer 20

Thanks. Good luck to you too.