Find $\lim_{k \to \infty} \frac{k^k}{(k+1)^{k+1}}$ I have spent the better part of this day trying to solve $$\lim_{k \to \infty} \frac{k^k}{(k+1)^{k+1}}$$ When I write this in Wolfram, it says the limit $= 0$, but I don't understand how it got to that conclusion.
 A: Hint for real numbers $(k+1)^{k+1}\geq k^k$ thus as the k becomes larger the denominator becomes so large that the whole fraction tends to $0$.
A: Recall that
$$\lim_{k \to \infty} \frac{k^k}{(k+1)^k} = \lim_{k \to \infty} \left( 1-\frac{1}{k+1} \right)^k = \lim_{k \to \infty} \left( \left( 1-\frac{1}{k+1} \right)^{k+1} \right)^{\frac{k}{k+1}} =\left( \frac{1}{e} \right)^1 = \frac{1}{e}$$
Now, your limit is
$$\lim_{k \to \infty} \frac{k^k}{(k+1)^k} \cdot \frac{1}{k+1} = \lim_{k \to \infty} \frac{k^k}{(k+1)^k} \cdot \lim_{k \to \infty} \frac{1}{k+1} = \frac{1}{e} \cdot 0 = 0$$
A: First, prove that :
$$ \lim \limits_{k \to +\infty} \Big( \frac{k}{k+1} \Big)^{k} = \frac{1}{e}. $$
Then, it follows that :
$$ \lim \limits_{k \to +\infty} \frac{k^{k}}{(k+1)^{k+1}} = 0 $$ 
because :
$$ \frac{k^{k}}{(k+1)^{k+1}} = \frac{1}{k+1} \Big( \frac{k}{k+1} \Big)^{k}. $$
A: $$\lim_{k \to \infty} \frac{k^k}{(k+1)^{k+1}}=\lim_{k \to \infty} \frac{k^k}{(k+1)^{k}}\cdot\frac{1}{k+1}=\lim_{k \to \infty} \left(\frac{1}{1+\frac1k}\right)^k\cdot\frac{1}{k+1}$$
$$\lim_{k \to \infty}\left(\frac{1}{1+\frac1k}\right)^k=e^{\lim_{k \to \infty}\left(\frac{1}{1+\frac1k}-1\right)\cdot k}=e^{\lim_{k \to \infty}\left(\frac{-1}{1+\frac1k}\right)}=e^{\lim_{k \to \infty}\left(\frac{-k}{1+k}\right)}=\frac1e$$
Thus,
$$\lim_{k \to \infty} \frac{k^k}{(k+1)^{k+1}}=\lim_{k \to \infty} \left(\frac{1}{1+\frac1k}\right)^k\cdot\lim_{k \to \infty} \frac{1}{k+1}=\frac1e\cdot0=0$$
A: For all $k > 1$,
$$ 0 < \frac{k}{k+1} < 1 $$
and 
$$ 0 < \left(\frac{k}{k+1}\right)^k < 1. $$
But
$$ \frac{k^k}{(k+1)^{k+1}} = \frac{1}{k+1} \left(\frac{k}{k+1}\right)^k. $$
Therefore
$$ 0 < \frac{k^k}{(k+1)^{k+1}} < \frac{1}{k+1} $$
for all $k>1$, and since $\lim_{k\to\infty} \frac{1}{k+1} = 0$,
by the squeeze theorem
$$ \lim_{k\to\infty} \frac{k^k}{(k+1)^{k+1}} = 0 $$
A: It's clear that $k^k$ is less than $(k+1)^k$. So you can make an upper bound of $\frac{1}{k+1}$ which clearly converges to zero. Hope this helps..
