Is the limit not infinity?

Question

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

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

Answer 1

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

Answer 2

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 :-)

Answer 3

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 $\lim\limits_{k\to +\infty} y=e$.

Answer 4

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

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

Answer 5

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 $\lim_{x\to -\infty} ln(1+\frac{1}{x}) \sim \frac{1}{x}$...