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Let $k \geq 6$ and I know $k!$ < $\dfrac{k^k}{2^k}$ I want to show the following:

$(k+1)! < \dfrac{(k+1)^{k+1}}{2^{k+1}}$

Now I am going to show my solution, let me know if my reasoning is correct as well.

Simplifying what we have above we get: $(k+1)k!$ < $\dfrac{(k+1)(k+1)^k}{2^{k+1}}$

Now divide $(k+1)$ from both sides: $k!$ < $\dfrac{(k+1)^k}{2^{k+1}}$

Next, factor out a k from the right side: $k!$ < $\dfrac{k^k(1+ \frac{1}{k})^k}{2^{k+1}}$

Now we know $\lim_{k\to\infty}$ $(1+ \frac{1}{k})^k$ is $e$ so we can write:

$k!$ < $\dfrac{k^ke}{2^{k+1}}$ $\iff$ $k!$ < $\dfrac{k^ke}{2\cdot2^k}$

This is also equivalent to: $k!$ < $\dfrac{k^k}{2^k}$ $\cdot$ $\dfrac{e}{2}$

Since I know $k!$ < $\dfrac{k^k}{2^k}$ (from an induction hypothesis) and I am multiplying by $\dfrac{e}{2}$ which is > $1$ the inequality holds. End of Proof.

The step I am most unsure of is setting the $(1+ \frac{1}{k})^k$ to $e$. Can I do this?

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  • $\begingroup$ No you can't. But you can probably easily show that $(1+1/k)^k\gt 2$. Then rewrite the proof so that you are not going backwards. $\endgroup$ – André Nicolas Sep 21 '12 at 2:14
  • $\begingroup$ Could you give me a hint about how to show that that statement is true? I have no idea. $\endgroup$ – CodeKingPlusPlus Sep 21 '12 at 2:29
  • $\begingroup$ Where does the $\dfrac{k(k-1)}{2}$ in front of $x^2$ come from? Shouldn't it be in front of $x$? And also in your example we get $\dfrac{3}{2}$ $\endgroup$ – CodeKingPlusPlus Sep 21 '12 at 5:07
  • $\begingroup$ Sorry, typo. For $k\ge 2$, $(1+x)^k\ge 1+kx+\frac{k(k-1)}{2}x^2$. (Shouldn't try to type math in comments.) Put $x=1/k$. The first two terms give 1+k(1/k)=2$, the next adds to it. Actually, you only need $\ge 2$, so $x^2$ term not needed. $\endgroup$ – André Nicolas Sep 21 '12 at 5:25
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You can use the inequality $$ \left(1 + \frac{1}{n}\right)^n < e < \left(1 + \frac{1}{n}\right)^{n+1}. $$ (This can be proved using the Taylor series of $\log$.)

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