Finding a Limit

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Hi,

I am at an intermedate step of a larger problem and I need to find the following limit:

lim(k->infinity) k^k/(k+1)^k

I see what this equals (by using computational software) but I am not sure how to show this.

Any help?
 

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  • #2
JasonRox
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mpm166 said:
Hi,

I am at an intermedate step of a larger problem and I need to find the following limit:

lim(k->infinity) k^k/(k+1)^k

I see what this equals (by using computational software) but I am not sure how to show this.

Any help?
Through the epsilon-delta definition of a limit?

Is the limit 0 or 1? I'm just taking a guess here without computing anything. Looks like 0 or 1 to me.

For advice, try writing it like this...

k^k/(k+1)^k = [k/(k+1)]^k

Look at the k and k+1.
 
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  • #3
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It's actually 1/e :smile:
 
  • #4
JasonRox
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Actually it's not 0 or 1, now that I looked at it for a minute.
 
  • #5
JasonRox
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mpm166 said:
It's actually 1/e :smile:
Yeah, I soon as I realized that it wasn't 0 or 1 it came to me that I saw it before.

Just didn't know where.
 
  • #6
nrqed
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mpm166 said:
Hi,

I am at an intermedate step of a larger problem and I need to find the following limit:

lim(k->infinity) k^k/(k+1)^k

I see what this equals (by using computational software) but I am not sure how to show this.

Any help?

Can be written as [itex] ({k \over k +1 })^k [/itex] which is [itex] ({ 1 + 1/k })^{-k} [/itex] and this is a well know form...As k goes to infinity it goes to [itex] e^{-1} [/itex]

(it is well known in the form [itex] lim_{n \rightarrow \infty} (1 + {1 \over n})^n = e [/itex])

Patrick
 
  • #7
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thanks for the help
 
  • #8
JasonRox
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nrqed said:
Can be written as [itex] ({k \over k +1 })^k [/itex] which is [itex] ({ 1 + 1/k })^{-k} [/itex] and this is a well know form...As k goes to infinity it goes to [itex] e^{-1} [/itex]

(it is well known in the form [itex] lim_{n \rightarrow \infty} (1 + {1 \over n})^n = e [/itex])

Patrick
How do you prove it though using the definition of a limit?

I'll give it a shot myself, but it doesn't seem like something that is pleasant to work with.
 
  • #9
nrqed
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JasonRox said:
How do you prove it though using the definition of a limit?

I'll give it a shot myself, but it doesn't seem like something that is pleasant to work with.
It`s probably not rigorous but one way is to consider instead the natural log of this expression. Setting x=1/n, one has to consider the limit as x goes to zero of ln(1+x)/x, an indeterminate form. Using L`Hospital rule, this gives 1. So the initial limit is e^1.

Patrick
 
  • #10
JasonRox
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nrqed said:
It`s probably not rigorous but one way is to consider instead the natural log of this expression. Setting x=1/n, one has to consider the limit as x goes to zero of ln(1+x)/x, an indeterminate form. Using L`Hospital rule, this gives 1. So the initial limit is e^1.

Patrick
Yeah, I see what you mean.

It's not rigorous in the sense that taking the log doesn't always work.

It would still be nice to see a proof though.
 
  • #11
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quick question, to solve this problem isn't this how you set it up?

[tex]1+e<k^2/(k^2+k)[/tex]
isn't the the right equation?
 
  • #12
JasonRox
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konartist said:
quick question, to solve this problem isn't this how you set it up?

[tex]1+e<k^2/(k^2+k)[/tex]
isn't the the right equation?
That's false though. Try k=1.

I think using the Squeeze Theorem might provide an answer though.
 
  • #13
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It would still be nice to see a proof though.
We can start from the series defintion of e

[tex]e = \sum \frac{1}{n!}[/tex]

Let [itex]S_n = 1 + 1 + \frac{1}{2!} + \cdots + \frac{1}{n!}[/tex] and [itex]T_n = (1 + \frac{1}{n})^n[/itex]. I'm gonna assume that T = lim T_n exists, although it can be proved by messing around with inequalities and the binomial theorem (if I recall correctly). Let's actually use the binomial theorem:

[tex]
\begin{align*}
T_n &= 1 + n\frac{1}{n} + \frac{n(n-1)}{2!}\frac{1}{n^2} + \cdots + \frac{n(n-1)(n-2)\cdots 1}{n!}\frac{1}{n^n} \\
&= 1 + 1 + \frac{1}{2!}(1 - \frac{1}{n}) + \cdots + \frac{1}{n!}(1 - \frac{1}{n})(1 - \frac{2}{n}) \cdots (1 - \frac{n-1}{n})
\end{align*}
[/tex]

Now notice that if N>n, then:

[tex]
T_N > 1 + 1 + \frac{1}{2!} (1 - \frac{1}{N}) + \cdots + \frac{1}{n!}(1 - \frac{1}{N}) \cdots (1 - \frac{n-1}{N})
[/tex]

Hence if we let N tend to infinity, the left side tends to T while the right side tends to S_n. Therefore we have S_n <= T. However we can also see that S_n >= T_n for all n. So we actually have T_n <= S_n <= T. And thus T = lim S_n = e.
 
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  • #14
JasonRox
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devious_ said:
We can start from the series defintion of e

[tex]e = \sum \frac{1}{n!}[/tex]

Let [itex]S_n = 1 + 1 + \frac{1}{2!} + \cdots + \frac{1}{n!}[/tex] and [itex]T_n = (1 + \frac{1}{n})^n[/itex]. I'm gonna assume that T = lim T_n exists, although it can be proved by messing around with inequalities and the binomial theorem (if I recall correctly). Let's actually use the binomial theorem:

[tex]
\begin{align*}
T_n &= 1 + n\frac{1}{n} + \frac{n(n-1)}{2!}\frac{1}{n^2} + \cdots + \frac{n(n-1)(n-2)\cdots 1}{n!}\frac{1}{n^n} \\
&= 1 + 1 + \frac{1}{2!}(1 - \frac{1}{n}) + \cdots + \frac{1}{n!}(1 - \frac{1}{n})(1 - \frac{2}{n}) \cdots (1 - \frac{n-1}{n})
\end{align*}
[/tex]

Now notice that if N>n, then:

[tex]
T_N > 1 + 1 + \frac{1}{2!} (1 - \frac{1}{N}) + \cdots + \frac{1}{n!}(1 - \frac{1}{N}) \cdots (1 - \frac{n-1}{N})
[/tex]

Hence if we let N tend to infinity, the left side tends to T while the right side tends to S_n. Therefore we have S_n <= T. However we can also see that S_n >= T_n for all n. So we actually have T_n <= S_n <= T. And thus T = lim S_n = e.
When you're taking N approaches infinity, are you assuming n is tagging along just behind it? Or is n fixed?
 
  • #15
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JasonRox said:
When you're taking N approaches infinity, are you assuming n is tagging along just behind it? Or is n fixed?
It's fixed of course. Maybe I should have stated that. :smile:
 
  • #16
JasonRox
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Ok, if N>n, then 1/N! < 1/n!. You can do this easily by induction.

But, you got 1/n! in your right hand side of that inequality. You replaced 1/N! with a bigger number 1/n!. Also, you replaced N with n for the numerators too, so that makes it smaller. Also, T_N has more terms than the right hand side.

After replacing certain N's with n and subtracting a few terms, how can you justify this inequality holds?

I'm not positive if I did this right, but T_2 = 2.25 so that's T_N. Now, N=2, so n=1 (only option), but on the right side of the inequality I'm getting 2.25.

Maybe it's not strictly larger? Or maybe you don't know the Latex code for it because you say "<=" later. :tongue2:

I'm going to have to write his out and see if I can satisfy myself. Looks right, but I'll just work out the details for myself.

The idea of the Squeeze Theorem is what I thought it would be.
 
  • #17
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No, I meant strictly larger.
 

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