The Darn e

[alexander]

So last week i was sitting in math class, bored out of my mind as usual, and a friend of mine throws me a problem and asks me to solve it to see if i can come up with the right solution, as he could not come up with it any way he tried

 

I find what the sequence converged to:

[math]a_n=\left( 1+\frac{2}{n}\right)^n[/math]

so you know, not thinking about it, jump right on that, try about a gazillion different ways to solve this, and what do i end up with? nothing remotely correct of an answer, i mean perhaps it was somewhat correct, but according to the book it was no such thing. it seems that every time i get to one, and that is just not right.

so this is basically what it constantly weilds down to something like this

[math]\lim_{n\rightarrow\infty}\left( 1+\frac{2}{n}\right)^n[/math]

so 1 is the same as n/n and adding n/n to 2/n weilds a n+2/n so i write

[math]\lim_{n\rightarrow\infty}\left(\frac{n+2}{n}\right)^n[/math]

then open parenthesees and i get

[math]\lim_{n\rightarrow\infty}\left(\frac{(n+2)^n}{n^n}\right)[/math]

so ok using propperties of logs

[math]\lim_{n\rightarrow\infty}\left(\frac{n\ln{(n+2)}}{n\ln{n}}\right)[/math]

so getting rid of the n, i divide through by n and get

[math]\lim_{n\rightarrow\infty}\left(\frac{\ln{(n+2)}}{\ln{n}}\right)[/math]

so then i apply the l'hospitals rule

[math]\lim_{n\rightarrow\infty}\left(\frac{d(\ln{n+2})dn}{d(\ln{n})dn}\right)[/math]

so that leads me to

[math]\lim_{n\rightarrow\infty}\left(\frac{\frac{1}{n+2}}{\frac{1}{n}}\right)[/math]

so outside over inside

[math]\lim_{n\rightarrow\infty}\left(\frac{n}{n+2} \right)[/math]

so l'hospitals rule again and i get a straight answer of one, uno. and about 3 steps ago i could see that this approach totally does not give me the right answer! so a few hours of that and it just hit me, like duh over my head!

now, the answer is [math]e^2[/math]

 

so this is the fun part for math wizes (aka people i am not)

 

(here is a small short contest, and it should be trivialy easy and quick to figure out) Post a solution to show why the answer is that (please include all the math that goes with it), i dont have much, but i can promiss a couple of positive rank points :)

 

This was such a brain teaser :P

[CraigD]

I had to find the sum of the series:

[math]\sum_{n=1}^\infty\left\lbrace 1+\frac{2}{n}\right\rbrace^n[/math]

I’m either misunderstanding the question, or this is a trick question.

 

[math]\sum_{n=1}^\infty 1 = 1 + 1 + 1 + 1 ... = \infty[/math]

 

[math]\left (1+\frac 2 n \right )^n > 1[/math] for all [math]n \ge 1[/math]

 

so ...

 

[math]\sum_{n=1}^\infty\left\lbrace 1+\frac{2}{n}\right\rbrace^n > \sum_{n=1}^\infty 1[/math]

 

so ...

 

[math]\sum_{n=1}^\infty\left\lbrace 1+\frac{2}{n}\right\rbrace^n = \infty[/math]

Q.E.D

 

[math]\lim_{n\rightarrow\infty}\left( 1+\frac{2}{n}\right)^n[/math]

is more interesting, but not necessary to answer the asked question.

[Jay-qu]

I dont see how you got that craig.. the first term would be (1+2/1)^1=3, second term (1+2/2)^2=9, third (1+2/3)^3=125/27 etc..

[ronthepon]

But it will be infinity anyway. I'll re-word what craig's put up.

 

[math] \frac{2}{n} [/math] will always be a positive number, so [math] 1 + \frac {2}{n} [/math] will always be more than one. That means that [math] \left( 1 + \frac {2}{n} \right)^n [/math] will also be over one. (limits are from one to positive infinity)

 

And adding an infinite number of positive terms, all greater than one will always yeild infinity.

 

er... right?

[CraigD]

… And adding an infinite number of positive terms, all greater than one will always yeild infinity.

 

er... right?

Right.

 

I think this question is an example of a common “trick question”, which asks an easy question, but contains information that can be used to ask a difficult one. The student is tempted to consider the difficult question before the stated one.

[Jay-qu]

Right.

 

I think this question is an example of a common “trick question”, which asks an easy question, but contains information that can be used to ask a difficult one. The student is tempted to consider the difficult question before the stated one.

There the best ones B)

 

Ron you are right, this is called this divergence.

[CraigD]

[math]\lim_{n\rightarrow\infty}\left( 1+\frac{2}{n}\right)^n[/math]

is more interesting …

We can’t just abandon the interesting question, can we? :)

 

As Alexander notes

so a few hours of that and it just hit me, like duh over my head!

now, the answer is [math]e^2[/math]

Exploring some examples suggest

 

[math]\lim_{n\rightarrow\infty}\left( 1+\frac{2}{n}\right)^n \overset?= e^2[/math]

 

A little more exploring suggests the more general

 

[math]\lim_{n\rightarrow\infty}\left( 1+\frac{k}{n}\right)^n \overset?= e^k[/math]

 

Proving this identity shows something remarkable about the constant [math]e[/math] and logarithms.

 

[math]\lim_{n\rightarrow\infty}\left( 1+\frac{k}{n}\right)^n \overset?= e^k[/math] [1]

 

[math]\lim_{n\rightarrow\infty} \ln \left (1+\frac{k}{n}\right)^n \overset?= \ln( e^k)[/math] [2]

 

[math]\lim_{n\rightarrow\infty} n \ln \left (1+\frac{k}{n}\right) \overset?= k[/math] [3]

 

Given the (remarkable) identity

 

[math]\lim_{x\rightarrow 0} \ln \left (1+x\right) = x[/math] [4]

 

which can be rewritten

 

[math]\lim_{n\rightarrow\infty} \ln \left (1+\frac{k}{n}\right) = \frac{k}{n}[/math] [5]

 

substituting [5] into [3], gives

 

[math]\lim_{n\rightarrow\infty} n \frac{k}{n} \overset?= k[/math] [6]

 

Completing the proof.

 

The remarkable identity

 

[math]\lim_{x\rightarrow 0} \ln \left (1+x\right) = x[/math]

 

is due to the identity

[math] \ln (y) )^\prime = \frac1y[/math]

 

It’s similar to the identity

 

[math]\lim_{\theta\rightarrow 0} \sin(\theta) = \theta[/math].

 

and means that, for [math]y[/math] values close to 1,

 

[math]\ln(y) \dot= y-1 [/math]

[Qfwfq]

This approximation is often used to make calculations more feasible.

[alexander]

i used the explore method too Craig, once i remembered that

[math]e=\left{1+\frac{1}{n}\right}^n[/math]

and later played with it and figured out the same thing you did craig

but other then playing with it there should be a way of getting there mathemagically!?!

 

Oh and my bad on the question part i got that totally wrong as i was really tired typing that (going on 36 hours without sleep and no coffee in site), so i will edit the top post to put up the right question... oopsie :)

[alexander]

really sorry for not checking that after i had some sleep... i really feel like a total dumb*$$ because i cound not get how i could post something like that and not check it... my bad guys!

[alexander]

hey you want an algebraic e brain teaser?

 

use the ratio test to test the convergence of the series:

[math]\sum_{n=1}^{\infty}\frac{n^n}{n!}[/math]

[Qfwfq]

Well, quite easily, it's divergent because n to the n beats the factorial. By the ratio criterion we get that the ratios not only are consistently > 1 but are even an increasing sequence, as the ratio trivially works out to:

 

[math](\frac{n+1}{n})^{n+1}[/math]

 

 

9 times 9 times 9... is bigger than 9 times 8 times 7...

[alexander]

and i thought you would show a shiny mathematical solution that has everything to do with e (the topic of this post).... :(

 

[math]\sum_{n=1}^\infty\frac{n^n}{n!}[/math]

the n and n+1 term

[math]a_n=\frac{n^n}{n!} a_{n+1}=\frac{(n+1)^{(n+1)}}{(n+1)!}[/math]

using the ratio test

[math]\frac{\frac{(n+1)^{(n+1)}}{(n+1)!}}{\frac{n^n}{n!}}[/math]

which is

[math]\frac{(n+1)^{(n+1)}*n!}{n^n*(n+1)!}[/math]

simplify more

[math]\frac{(n+1)^n*\sout{(n+1)*n!}}{n^n*\sout{(n+1)*n!}}[/math]

 

[math]\left(\frac{n+1}{n}\right)^n[/math]

limit of that is

[math]\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e[/math]

 

since the value is more then one, sequence is divergent

 

i just thought that it was weird that a similar limit answer came up so quickly, i rarely see something like that, ever, infact once or twice in the past 2 years, and here they are back to back within a week... weird

[Qfwfq]

Sure, I hadn't noticed the limit being e and my quick scribbles weren't even exact :singer: but isn't it a shiny enough mathematical solution to point out that the series lacks a fundamental requisite for being convergent? When the n-th addend isn't infinitesimal the series can't be convergent. When it is increasing as in this case, the series can only be divergent. By Occam's razor, neither the ratio nor other criteria were needed.

[alexander]

ofcourse it was Q, at least your way of doing this was way better then my teacher doing a somewhat similar problem, he used the "Oh C'mon" theorem to prove that [math]\lim_{\theta\rightarrow0} \frac{sin\theta}{\theta}=1[/math] a while back

 

but as i said it was just weird to get e there :)

[Qfwfq]

Gee, I had never heard of the "Oh C'mon" theorem :D it must be a great one! Could you kindly state it and demonstrate?

 

I do know the "2 Cops" theorem though:

 

Hyp:

 

[math]f(x)\le h(x)\le g(x)[/math]

 

and

 

[math]\lim_{x\rightarrow p}f(x)=\lim_{x\rightarrow p}g(x)=a[/math]

 

Thes:

 

[math]\lim_{x\rightarrow p}h(x)=a[/math]

[alexander]

i will demonstrate the oh c'mon theorem, but doing this just for you, Q, just so you know :)

 

the Oh C'mon theorem states: "oh, cooome ooon"

 

basically, this a a typica proof:

prove that [math]\lim_{x\rightarrow0} \frac{sin x}{x} = 1[/math]

first assume that theta lies between 0 and [math]\frac{\pi}{2}[/math] on a unit circle. Triangle ABC is drawn from the center of the circle A to a point on the circomference B at the angle x. Height BD is drawn perpendicular to AC. From this we get that |BC| < |BC| < arc BC. Therefore sin x < x thus sin x / x < 1.

Let the tangent lines of BC intersect at E and AB is extended to CE and intersects at point F. (shesh i wish i had a picture)

This makes the arc BC inscribed in a triangle BCE. Thus x=arc BC < |CE| + |EB| < |CE| + |EF| = |CF| = |AC| tan x = tan x

Thus we have [math]x < \frac{sin x}{cos x} [/math] so [math]cos x < \frac{sin x}{x} < 1[/math]

We know that [math]\lim_{x\rightarrow0} 1 = 1[/math] and that [math]\lim_{x\rightarrow0} cos x = 1[/math]

so by the Squeeze Theorem we have [math]\lim_{x\rightarrow0^+} \frac{sin x}{x} = 1[/math]

but the function sin(x)/x is even so its right and left limits are equal, thus the theorem is proven.

 

yes i could go into the proof of the squeeze theorem and prove that the function is even, but no time nor want to do that, the above is pleny enough for the moment.

 

So here is the contrast proof using the Oh C'mon theorem:

[math]lim_{x\rightarrow0} \frac{sin x}{x} =^h \frac{cos x}{1} = 1[/math]

explained: using the l'hospitals rule (i denoted by the h over the = sign), oh c'mon we dont have the time for a real proof, we can find thy limit by taking a derivative of the top and the bottom constants, then substituting one for the limit of cos x and we get 1/1 or one.