Factoring out an exponential?

Question

I have the following expression

$$\frac{2^{k+1}(k+1)!}{(k+1)^{k+1}}\cdot\frac{k^k}{2^k k!}$$

I get

$$\frac{2(k+1)(k^k)}{(k+1)^{k+1}}$$

But how do I factor out the ${(k+1)}^{k+1}$

Answer 1

It might help if you notice that $(k+1)^{k+1}=(k+1)^k(k+1)$.

Answer 2

You have

$$\frac{2k^k(k+1)}{(k+1)^{k+1}}=\frac{2k^k}{(k+1)^k}=2\left(\frac{k}{k+1}\right)^k=2\left(1-\frac1{k+1}\right)^k\;;$$

that last form should be easier to work with if you want the limit as $k\to\infty$.

Answer 3

$$\frac{2^{k+1}(k+1)!}{(k+1)^{k+1}}\cdot\frac{k^k}{2^k k!}=$$ $$=\frac{2^{k}2(k+1)k!}{(k+1)^{k}(k+1)}\cdot\frac{k^k}{2^k k!}=$$

$$=\frac{2k^k}{(k+1)^k}=2\left(\frac{k}{k+1}\right)^k$$