Question

why

Answer 1

is \[\lim_{k \rightarrow \infty}\frac{ k^k }{ (k+1)^k} = \frac{ 1 }{ e } \]

Answer 2

Yes

Answer 3

You can prove this. Should I start you off?

Answer 4

please

Answer 5

\(\displaystyle \lim_{k\to \infty}\frac{k^k}{(k+a)^k}=\lim_{k\to \infty}\left[\frac{k}{k+a}\right]^k=\lim_{t\to \infty}\left[\frac{t-a}{t}\right]^t\)

You can achieve the third step in that line by setting k+1=t (that dictates k=t-1).
(If you have a question about that, ask...)
Then use \(\displaystyle y=e^{\ln y}\).

\(\displaystyle \lim_{t\to \infty}\left[\frac{t-a}{t}\right]^t=\exp\left\{\ln~\lim_{t\to \infty}\left[\frac{t-a}{t}\right]^t\right\}\)
\(\displaystyle =\lim_{t\to \infty}e^{\ln~\left[\frac{t-a}{t}\right]^t}=\lim_{t\to \infty}e^{t\ln~\left[\frac{t-a}{t}\right]}=\lim_{t\to \infty}e^{\ln~\left[\frac{t-a}{t}\right]~/~(1/t)}\)

In the last step I re-wrote t as 1/t in the denominator.
Now, take the L'Hospital's Rule.

Answer 6

By the way, `exp{y}` is the same as `e^y`

Answer 7

You can use 1, instead of "a", if abstractness troubles you. (B/c in your case you have 1, not a.) I was just going in general. \(\\[0.9em]\)
[In my general proof startup, you should get e^(-a).]

Answer 8

One more thing. It would reduce paint, if you re-write `(t-a)/t` as `1 - (a/t)`.

Answer 9

still here, working through it...

Answer 10

Yes, just to show where we are, you are to take the L'H'S of
\(\displaystyle \exp\left\{\lim_{t\to \infty}~\frac{\ln\left[1-\frac{a}{t}\right]}{1/t}\right\}\)

Answer 11

you can ask me to repost it but without confusions. I will do so, if you need.

Answer 12

so, \(\color{blue}{\text{Originally Posted by}}\) @SolomonZelman
\(\displaystyle \lim_{k\to \infty}\frac{k^k}{(k+a)^k}=\lim_{k\to \infty}\left[\frac{k}{k+a}\right]^k=\lim_{t\to \infty}\left[\frac{t-a}{t}\right]^t\)

You can achieve the third step in that line by setting k+1=t (that dictates k=t-1).
(If you have a question about that, ask...)
Then use \(\displaystyle y=e^{\ln y}\).

\(\displaystyle \lim_{t\to \infty}\left[\frac{t-a}{t}\right]^t=\exp\left\{\ln~\lim_{t\to \infty}\left[\frac{t-a}{t}\right]^t\right\}\)
\(\displaystyle =\lim_{t\to \infty}e^{\ln~\left[\frac{t-a}{t}\right]^t}=\lim_{t\to \infty}e^{t\ln~\left[\frac{t-a}{t}\right]}=\color{red}\lim_{t\to \infty}e^{\ln~\left[\frac{t-a}{t}\right]~/~(1/t)}\)

In the last step I re-wrote t as 1/t in the denominator.
Now, take the L'Hospital's Rule.
\(\color{blue}{\text{End of Quote}}\)

So the limit in red is where Im having some trouble

Answer 13

Where did the 1/t come from?

Answer 14

Oh.

t = 1/ (1/t)

Answer 15

Multiply by t, it same as dividing by 1/t, isn't it?

Answer 16

or, do you mean why I did that?

Answer 17

so you could apply L'H?

Answer 18

Yes, because as t->∞, the following is true:

1) ln[ t-a / a ] = ln[1] = 0
2) 1/t = 1/∞ .... so to speak .... and that is =0.

So we have a classic 0/0 case

Answer 19

oh, (1) should be t-a / t

Answer 20

sorry:)

Answer 21

I am doing a complete copy of the proof in a different thread.
If you want you can as well wait for that ...

Answer 22

Sure! Sorry, hopping between this and lecture right now. Final is next week so going a little crazy

Answer 23

I got part of it down ... I will post I got so far.

\(\color{blue}{\displaystyle \lim_{k\to \infty}\frac{k^k}{(k+\beta )^k}=\lim_{k\to \infty}\left(\frac{k}{k+\beta}\right)^k }\)
\(\tiny \\[0.9em]\)
I will make a substitution: \(\displaystyle k+\beta=t \)
This condition dictates: \(\displaystyle k=t-\beta \)
And when \(\displaystyle k\to \infty \), I know that \(\displaystyle t\to \infty \) as well. \(\tiny \\[0.7em]\)

\(\color{blue}{\displaystyle \lim_{k\to \infty}\left(\frac{k}{k+\beta}\right)^k==\lim_{t\to \infty}\left(\frac{t-\beta}{t}\right)^{t-1} }\)
\(\tiny \\[1.3em]\)
Now, I will reduce this limit:
\(\color{blue}{\displaystyle \lim_{t\to \infty}\left(\frac{t-\beta}{t}\right)^{t-1}=\lim_{t\to \infty}\left(\frac{t-\beta}{t}\right)^{t}\times \lim_{t\to \infty}\left(\frac{t-\beta}{t}\right)^{-1} }\) \(\tiny \\[1.7em]\)
\(\color{blue}{\displaystyle =\lim_{t\to \infty}\left(\frac{t-\beta}{t}\right)^{t}\times \lim_{t\to \infty}\left(\frac{t}{t-\beta}\right)^{1} }\)
\(\color{blue}{\displaystyle =\lim_{t\to \infty}\left(\frac{t-\beta}{t}\right)^{t}\times \color{red}{1} }\)

So, so far I know the following equivalence:
\(\color{black}{\displaystyle \lim_{k\to \infty}\frac{k^k}{(k+\beta )^k}:=\lim_{t\to \infty}\left(\frac{t-\beta}{t}\right)^t }\)
I will proceed to evaluate the last limit (to prove the initial one).

I will re-write the last limit as follows (using log. propreties):
\(\color{black}{\large \displaystyle e^{\lim_{t\to \infty}~\ln \left[\left(\frac{t-\beta}{t}\right)^t\right] } }\)


and to avoid writing exponent of e every time, will just examine
\(\color{black}{ \displaystyle \lim_{t\to \infty}~\ln \left[\left(\frac{t-\beta}{t}\right)^t\right] }\).

\(\color{blue}{ \displaystyle \lim_{t\to \infty}~\ln \left[\left(\frac{t-\beta}{t}\right)^t\right] =\lim_{t\to \infty}~t\cdot \ln \left[\frac{t-\beta}{t}\right]=\lim_{t\to \infty}~\frac{\ln \left[\frac{t-\beta}{t}\right] }{1/t} }\)

Answer 24

this is where we got till, take your time to read if you need .... (sorry for making it long, tried to make it readable and simple)

Answer 25

\(\color{blue}{ \displaystyle \lim_{t\to \infty}~\frac{\ln \left[\frac{t-\beta}{t}\right] }{1/t}=\lim_{t\to \infty}~\frac{\ln \left[1-\frac{\beta}{t}\right] }{1/t}=\lim_{t\to \infty}~\frac{\frac{d}{dt}\left\{\ln \left[1-\frac{\beta}{t}\right]\right\} }{\frac{d}{dt}\left\{1/t\right\} } }\)

Applying the rule for the derivative of ln(x), and the chain rule for inner argument, which is a power rule ... for the bottom, just the power rule

\(\color{blue}{ \displaystyle=\lim_{t\to \infty}~\frac{\frac{1}{1-\frac{\beta}{t} }\times \frac{\beta }{t^2} }{-1/t^2} }\)

Answer 26

now, what is needed is just to simply this limit.

You can cancel 1/t^2 on top and bottom.
\(\color{blue}{ \displaystyle=\lim_{t\to \infty}~\frac{\frac{\beta}{1-\frac{\beta}{t} } }{-1} =-\beta\cdot \lim_{t\to \infty}~\frac{1}{1-\frac{\beta}{t} } }\)

\(\color{blue}{ \displaystyle=-\beta\cdot \lim_{t\to \infty}~\frac{1}{\frac{t-\beta}{t} }=-\beta\cdot \lim_{t\to \infty}~\frac{t}{t-\beta } =-\beta\times \color{red}{1} }\)

Answer 27

Came to an end, just need to put the pieces to get:

That limit that I examined (which was supposed to be in the exponent of e) is equal to \(-\beta\), therefore the whole limit is \(e^{-\beta}\). And thus you know that

\(\color{blue}{\displaystyle \lim_{k\to \infty}\frac{k^k}{(k+\beta )^k}=e^{-\beta} }\)

Answer 28

It is actually not that hard, I just wrote it in a very lengthy way ... if you have questions, ask.

Answer 29

Well, if you really need something very short, then you can remember that

\(\displaystyle \lim_{x \rightarrow \infty}\left[\frac{ x+a }{ x}\right]^x = e^a\)

(your initial limit is also a version of this, just that a=-1 in your particular case, and to now this, you can just set a sub x+a=t, like I did in my proof.)

Answer 30

think this depends largely on your definition of \(e\)

Answer 31

I feel like there's a far easier/shorter method here.

First remember the definition of e
http://output.to/sideway/images/knowledge/mathematics/calculus/001/limit_a_01j.png
Do some simplifying
\[\large \lim_{k \rightarrow \infty}\frac{ k^k }{ (k+1)^k} = \] \[\large \lim_{k \rightarrow \infty}\left( \frac{ k }{ k+1} \right)^k=\] \[\large \lim_{k \rightarrow \infty}\left( \frac{ k+1 }{ k} \right)^{-k}=\] \[\large \lim_{k \rightarrow \infty}\left( 1+\frac{ 1 }{ k} \right)^{-k}=\]apply the power limit law seen here
http://image.slidesharecdn.com/lecture5-limitlaws-140916111716-phpapp01/95/lecture-5-limit-laws-7-638.jpg?cb=1410866284 \[\large \left( \lim_{k \rightarrow \infty}\left( 1+\frac{ 1 }{ k} \right)^{k} \right)^{-1}=\]Now look back at the definition of e.

Answer 32

\[\large \left( \lim_{k \rightarrow \infty}\left( 1+\frac{ 1 }{ k} \right)^{k} \right)^{-1}= e^{-1}\]

Answer 33

guess @agent0smith beat me to it.
if that is your definition of \(e\) then you are done in basically two steps

Answer 34

Ah, I see. Very cool. I like both ways, Soloman's way really showed me how it works and Agent show'd me the rule you can use to also verify it. Thanks for the detailed and well thought out responses guys, it's really helpful.