Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is the limit of this not infinity? No matter what the value of p is? Or is there a way to simplify that fraction?

$$\lim_{k \to \infty} 2^{p}\left(\frac{k}{k+1}\right)^k$$

share|improve this question
4  
Is that $2^p or 2^k$? If it is p as it looks you can remove that as it is a constant. –  Belov Apr 19 '13 at 18:43
3  
Answer to title question: "it's not." –  rschwieb Apr 19 '13 at 18:58
1  
Actually, the answer "it's not" would imply that the term goes to $\infty$ ;-) –  andreas Apr 19 '13 at 22:36
2  
There is no such limit as infinity, which is not a number. A limit expression either converges on a value (a limit exists and is that value) or else there does not exist a limit: as the limit-generating parameter increases or decreases, the value oscillates without settling, or grows without bound. –  Kaz Apr 19 '13 at 22:52
1  
@Kaz: It's pretty common to say that if for any positive number $M$ there's an $n$ such that from then on the sequence stays above $M$, then the limit is infinity. –  Javier Badia Apr 20 '13 at 0:13

5 Answers 5

Hint: $\displaystyle \lim \limits_{k\to +\infty} 2^p\left(\frac{k}{k+1}\right)^k=2^p\lim \limits_{k\to +\infty} \left(\frac{k}{k+1}\right)^k=2^p \left[\lim \limits_{k\to +\infty} \left(\frac{k+1}{k}\right)^k\right]^{-1}$

share|improve this answer
2  
The last equality is a bit dangerous if you don't know anything about the limit (just to make sure the thread starter knows that). –  andreas Apr 19 '13 at 22:41
1  
@andreas I agree, but since I meant to give a hint and not a full answer, I chose to answer this way. –  Git Gud Apr 20 '13 at 0:37

The limit involves the quantity $k$, while the term $2^p$ does not. So you can take that out of the limit

$$ 2^p \lim_{k\rightarrow\infty} \left(\frac{k}{k+1}\right)^k $$

That limit should begin to look familiar. In particular, we know that Euler's constant $e$, is given by:

$$ e = \lim_{k\rightarrow\infty} \left(1+\frac{1}{k}\right)^k $$

You can work with your limit to try and get it in that form and see if an $e$ pops our somewhere. Hint, it does :-)

share|improve this answer

Moving the constant out, you only have to deal with $\lim\limits_{k\to +\infty}\left(\frac{k}{k+1}\right)^k$. This is a $1^\infty$ indeterminate form, which you can analyze by setting $y=\left(\frac{k}{k+1}\right)^k$ and then examining $\ln y=k\ln\left(\frac{k}{k+1}\right)=\frac{\ln\left(\frac{k}{k+1}\right)}{\frac{1}{k}}$.

Using L'Hopital you get that $\lim\limits_{k\to +\infty}\ln y=1$ and so $\lim\limits_{k\to +\infty} y=e$.

share|improve this answer
    
But now seeing all of the other answers, it is definitely more efficient to shortcut with recognizing the limit for $e$ :) –  rschwieb Apr 19 '13 at 18:57
    
Good edits, much better than not recognizing the $1^\infty$ problem –  rajb245 Apr 19 '13 at 19:32

Use $\frac{k}{k+1} = \frac{k+1-1}{k+1} = 1 - \frac{1}{k+1}$ and $e^x = \lim_{k \leftarrow \infty} (1+\frac{x}{k})^k$. Clearly you get

$$\lim_{k \rightarrow \infty} 2^p \left(\frac{k}{k+1}\right)^k = 2^p \lim_{k \rightarrow \infty}\left(1 - \frac{1}{k+1}\right)^k = 2^p \lim_{k \rightarrow \infty}\left(1 + \frac{-1}{k}\right)^{k-1} = 2^p e^{-1} = 2^p/e$$

share|improve this answer

You can notice that

$$(\frac{k}{k+1})^k = \exp(k\ln(\frac{k}{k+1})) = \exp(k\ln(1-\frac{1}{k+1}))$$

Now if you remember that $\lim_{x\to -\infty} ln(1+\frac{1}{x}) \sim \frac{1}{x}$...

share|improve this answer
    
Nice approach, but there should be a - within your log –  andreas Apr 19 '13 at 22:34
    
You're right ! Thanks, I edited it. –  Robin Apr 19 '13 at 22:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.