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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 ${(k+1)}^{k+1}$

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  • $\begingroup$ Are you missing parentheses around the $k+1$? $\endgroup$ Nov 6, 2013 at 17:18
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    $\begingroup$ You don’t, apart from cancelling one factor of $k+1$. What are you trying to do with this expression? Take its limit as $k\to\infty$? $\endgroup$ Nov 6, 2013 at 17:22
  • $\begingroup$ yes I am trying to take the limit as k approach infinity $\endgroup$ Nov 6, 2013 at 17:29

3 Answers 3

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

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  • $\begingroup$ how does such a factoring work I am wondering $\endgroup$ Nov 6, 2013 at 17:21
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    $\begingroup$ @FernandoMartinez Think about what $a^k$ means. $a^2 = a \times a$, $a^3 = a \times a \times a$, etc. So $a^3 = a^2 \times a$. $\endgroup$ Nov 6, 2013 at 17:23
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    $\begingroup$ @FernandoMartinez - It is the standard definition of exponents, i.e. $a^{b}\cdot a^c=a^{b+c}$. Therefore $a^k\cdot a=a^{k+1}$. Now let $a=k+1$ and we get: $(k+1)^k\cdot (k+1)=(k+1)^{k+1}$ $\endgroup$
    – Mufasa
    Nov 6, 2013 at 17:24
  • $\begingroup$ I see yes it makes sense $\endgroup$ Nov 6, 2013 at 17:29
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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 $k\to\infty$.

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  • $\begingroup$ I think $(\frac{k}{k+1})^k$ is $\frac{1}{e}$ $\endgroup$ Nov 6, 2013 at 17:36
  • $\begingroup$ So my limit would be 2/e $\endgroup$ Nov 6, 2013 at 17:37
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    $\begingroup$ @Fernando: Looks good to me. $\endgroup$ Nov 7, 2013 at 9:53
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$$\frac{2^{k+1}(k+1)!}{(k+1)^{k+1}}\cdot\frac{k^k}{2^k k!}=$$ $$=\frac{2^{k}2(k+1)k!}{(k+1)^{k}(k+1)}\cdot\frac{k^k}{2^k k!}=$$

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

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