Euler's number e, proving convergence and bounds

[Nerpilis]

The transcendental number e – Euler’s number
(the underscores(_) represent subscript and (^) represent superscript/exponet)
The limiting value of sequence {e_n} where e_n = (1+1/n)^n is the irrational number e. My text gives a challenge to see if you can prove that this converges by verifying the following: (1+1/n) ^n < e < (1+1/n) ^(n+1) and that 2 ≤ e < 3 for all n Є N. There has been a consensus in the class that there is a typo in the text and that it should read: (1+1/n) ^n < e_(n+1) < (1+1/n)^( n+1) 2 ≤ e_m < 3. I’m not sure if I know enough to question this but I will accept it and proceed with the hints on how to do this.

The first hint is to
Use the binominal theorem to write out n+1 terms of e_n using the fact that n(n-1)(n-2)/n^3 can be expressed as (1-1/n)(1-2/n). then do the same for n+2 in terms of e_(n+1) such that you can come to the conclusion that e_n < e_(n+1) right off the bat in this multi step problem I’m stumped am I using the binominal expansion correctly?. So far I’ve come up with:

e_n = 1 + 1 + (1/2)(n)(n-1)(1/n^2) + (1/3)(n)(n-1)(n-2)(1/n^3) + …. + (1/n)^n
e_n = 2 + (1/2)(n)(n-1)(1/n^2) + (1/3)(1-(1/n))(1-(2/n)) + …. + (1/n)^n

then for n+1 in terms of e_n

= 2 + (1/2)(n+1)(n)(1/(n+1)^2) + (1/3)(1-(1/(n+1))(1-2/(n+1)) + …. + (1/(n+1))^(n+1)

Then for n+2 in terms of e_(n+1) here I substituted n+2 for n in the equation above, is this correct according to the given hints?

= 2 + (1/2)(n+3)(n+2)(1/(n+3)^2) + (1/3)(1-(1/(n+3))(1-2/(n+3)) + …. + (1/(n+3))^(n+3)

I wanted to get some help on these preliminary steps before I move onto the second half of the challenge. If I am correct so far how does this come to the conclusion that e_n < e_(n+1)?

 

[VietDao29]

Oh, it's very hard to read... You can learn how to LaTeX at the General Physics board.

e_n = 1 + 1 + (1/2)(n)(n-1)(1/n^2) + (1/3)(n)(n-1)(n-2)(1/n^3) + …. + (1/n)^n
e_n = 2 + (1/2)(n)(n-1)(1/n^2) + (1/3)(1-(1/n))(1-(2/n)) + …. + (1/n)^n

then for n+1 in terms of e_n

= 2 + (1/2)(n+1)(n)(1/(n+1)^2) + (1/3)(1-(1/(n+1))(1-2/(n+1)) + …. + (1/(n+1))^(n+1)

Then for n+2 in terms of e_(n+1) here I substituted n+2 for n in the equation above, is this correct according to the given hints?

= 2 + (1/2)(n+3)(n+2)(1/(n+3)^2) + (1/3)(1-(1/(n+3))(1-2/(n+3)) + …. + (1/(n+3))^(n+3)

Nope that's not quite correct, you cannot substitute n + 1 or n + 2 for n, note that e_n has n + 1 terms while e_{n + 1} has n + 2 terms, and e_{n + 2} has n + 3 terms. You should again use binomial theorwm to write out the terms for e_{n + 1}, and e_{n + 2}

You seem to forget the factorial sign in while expanding the terms out ...
It should read something like:
(a + b) ^ n = a ^ n + na ^ {n - 1}b + \frac{n(n - 1)}{2}a ^ {n - 2}b ^ 2 + \frac{n(n - 1)(n - 2)}{3!}a ^ {n - 3}b ^ 3 + ... + b ^ n

It seem that you forget to rewrite your third terms for e_n. It can be written as:
\frac{1}{2} \times \frac{n(n - 1)}{n ^ 2} = \frac{1}{2} \times \frac{n}{n} \times \frac{n - 1}{n} = \frac{1}{2} \times \left( 1 - \frac{1}{n} \right)

Then you can continue your proof by noting that:
\frac{m}{n} &gt; \frac{m}{n + 1}, \forall n \geq 1, m &gt; 0
Hence
1 - \frac{m}{n} &lt; 1 - \frac{m}{n + 1}, \forall n \geq 1, m &gt; 0

Can you go from here?
Viet Dao,

 

[Nerpilis]

Thanks for the latex info, i was wondering how to enact the various notations.
Before I continue with the proof I want to verify that I'm using the binominal theorem correctly as well as latex.

e_{n+1}=2+\frac{n-1}{2n}+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+...+(\frac{1}{n})^n+(\frac{1}{n+1})^{n+1}

e_{n+2}=2+\frac{n-1}{2n}+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+...+(\frac{1}{n})^n+(\frac{1}{n+1})^{n+1}+(\frac{1}{n+2}^{n+2})I feel like I'm missing something, but anyway if I'm correct on the above expansion it seems that it would follow that e_{n}&lt;e_{n+1} since there is an additional expansion term that is >0 for all n\in{N}

I do follow your the latter part of how to do the proof, however my text also throws in that i should use this additional inequality n^{n}\geq{n!}\geq{2}^{n-1} to prove that e_{n}&lt;3. if I get that, from here i can use the squeez theroem to prove that e is bounded between 2 and 3. I guess I should accept the given inequality to be true, but i thought that factorials grow larger faster than exponets. Thanks again for the help

 

[VietDao29]

...I want to verify that I'm using the binominal theorem correctly as well as latex.

e_{n+1}=2+\frac{n-1}{2n}+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+...+(\frac{1}{n})^n+(\frac{1}{n+1})^{n+1}

e_{n+2}=2+\frac{n-1}{2n}+\frac{1}{3!}(1-\frac{1}{n})(1-\frac{2}{n})+...+(\frac{1}{n})^n+(\frac{1}{n+1})^{n+1}+(\frac{1}{n+2}^{n+2})I feel like I'm missing something...

Yes, you do miss something. I think you should check the expansion for e_{n + 1}, and e_{n + 2} again:
e_{n + 1} = 1 + \frac{n + 1}{n + 1} + \frac{1}{2} \times \frac{(n + 1)n}{(n + 1) ^ 2} + \frac{1}{3!} \times \frac{(n + 1)n(n - 1)}{(n + 1) ^ 3} + ... + \frac{1}{(n + 1)!} \times \frac{(n + 1)n(n - 1) ... 1}{(n + 1) ^ {n + 1}}
= 2 + \frac{1}{2} \times \frac{n}{(n + 1)} + \frac{1}{3!} \times \frac{n(n - 1)}{(n + 1) ^ 2} + ... + \frac{1}{(n + 1)!} \times \frac{n(n - 1) ... 1}{(n + 1) ^ {n}}

= 2 + \frac{1}{2} \times \left( \frac{n + 1 - 1}{(n + 1)} \right) + \frac{1}{3!} \left( \frac{(n + 1 - 1)(n + 1 - 2)}{(n + 1) ^ 2} \right)
+ ... + \frac{1}{(n + 1)!} \times \frac{(n + 1 - 1)(n + 1 - 2)(n - + 1 - 3)...(n + 1 - n)}{(n + 1) ^ {n}}

= 2 + \frac{1}{2} \times \left( 1 - \frac{1}{(n + 1)} \right) + \frac{1}{3!} \left( 1 - \frac{1}{(n + 1)} \right) \times \left( 1 - \frac{2}{(n + 1)} \right) + ... +
+ \frac{1}{(n + 1)!} \left( 1 - \frac{1}{n + 1} \right) \left( 1 - \frac{2}{n + 1} \right) \left( 1 - \frac{3}{n + 1} \right)...\left( 1 - \frac{n}{n + 1} \right)
Now, you can use the last part of my previous post to prove that e_n < e_{n + 1}

I do follow your the latter part of how to do the proof, however my text also throws in that i should use this additional inequality n^{n}\geq{n!}\geq{2}^{n-1} to prove that e_{n}&lt;3. if I get that, from here i can use the squeez theroem to prove that e is bounded between 2 and 3. I guess I should accept the given inequality to be true, but i thought that factorials grow larger faster than exponets. Thanks again for the help

Of course, it's obvious that e_n > 2.
You can also prove the inequality n! >= 2^{n - 1}:
n! = 1 . 2 . 3 . 4 . 5 . ... . n = 2 . 3 . 4 . ... . n >= 2 . 2 . 2 . 2 . ... . 2 = 2^{n - 1} Q.E.D
To continue your proof, you should note that:
1 - \frac{1}{m} &lt; 1 , \forall m &gt; 0
and \frac{1}{n!} \leq \frac{1}{2 ^ {n - 1}} , \forall n &gt; 0
Can you go from here?
Viet Dao,

 

[Nerpilis]

wow... I think I need further instruction on the binominal expansion. I'm a bit confused (not in the algebra afterwards, but how the part with the 3! term:
\frac{1}{3!} \times \frac{(n + 1)n(n - 1)}{(n + 1) ^ 3}

this is probably due to my misunderstanding of this theorem but why is there (n+1)! in the last term
\frac{1}{(n + 1)!} \times \frac{(n + 1)n(n - 1) ... 1}{(n + 1) ^ {n + 1}}

when in the e_{n} last term there isn't the factorial?

 

[VietDao29]

You should re-read your book on binomial expansion. Or you can also read it here (pay attention to the I part).
It says:
(x + y) ^ n = \sum_{k = 0} ^ n \left( \begin{array}{l} n &amp; k \end{array} \right) x ^ k y ^ {n - k}
Where:
\left( \begin{array}{l} n &amp; k \end{array} \right) = \frac{n!}{k!(n - k)!} = \frac{1}{k!} \times (n - k + 1)(n - k + 2) ... n
So let x = 1 / n, and y = 1.
The forth term of the expansion for e_{n + 1} will be:
\left( \begin{array}{c} n + 1 &amp; 3 \end{array} \right) \frac{1}{(n + 1) ^ 3} \times 1 ^ {n + 1 - 3} = \frac{(n + 1)!}{3!(n - 2)!} \times \frac{1}{(n + 1) ^ 3} \times 1 ^ {n - 2}
= \frac{1}{3!} (n - 1)n(n + 1) \times \frac{1}{(n + 1) ^ 3} \times 1 ^ {n + 1 - 3} = \frac{1}{3!} \times \frac{(n + 1)n(n - 1)}{(n + 1) ^ 3}.
So do you get it now?
For the last term, you may write it as:
\frac{1}{(n + 1) ^ {n + 1}}
But writing it as:
\frac{1}{(n + 1)!} \times \frac{(n + 1)n(n - 1) ... 1}{(n + 1) ^ {n + 1}} makes it easier for you to prove e_{n + 1} > e_n
They are the same:
\frac{1}{(n + 1)!} \times \frac{(n + 1)n(n - 1) ... 1}{(n + 1) ^ {n + 1}} = \frac{1}{(n + 1)!} \times \frac{(n + 1)!}{(n + 1) ^ {n + 1}} = \frac{1}{(n + 1) ^ {n + 1}}.
Viet Dao,