Find the maximum value of $\frac{101^{k/2}}{k!}$ where $k$ is a positive integer

Question

The positive integer $k$ for which $\dfrac{101^{k/2}}{k!}$ is maximum.
A) $9$
B) $10$
C) $11$
D) $101$

I have no idea how to solve this problem.

At first, I thought of taking this a function and then finding the roots of its derivative. \begin{align*} y &= \frac{101^{x/2}}{x!} \\ \ln y &= \left(\frac{x}{2}\right)\ln(101) - [\ln x + \ln(x-1) + \ln(x-2) +\cdots+ \ln2]. \end{align*}

But then, how do you differentiate $(\ln k + \ln(k-1) + \ln(k-2) +\cdots+ \ln2)$

Any help would be appreciated.

Answer 1

Taking finite differences will help here. When moving from $k$ to $k+1$:

• the numerator multiplies by $\sqrt{101}=10\dots$
• the denominator multiplies by $k+1$

Thus if $\sqrt{101}>k+1$, the expression increases going from $k$ to $k+1$, and the largest $k$ for which this is true is $k=9$. From $k=10$ to $k=11$ the expression decreases, so the maximum is attained at $k=10$.

Answer 2

You have been given simple and elegant solutions to your problem.

You could have obtained a result using your approach with the derivative considering the maximization of

$$\ln y = \left(\frac{x}{2}\right)\ln(101) - \log( x!)$$ and use Stirling approximation $$\log(x!)=x (\log (x)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({x}\right)\right)+O\left(\frac{1}{x}\right)$$ making $$\frac {y'}y=\frac{-2 x \log (x)+x \log (101)-1}{2 x}\implies 2 x \log (x)-x \log (101)+1=0$$ Neglecting the $1$ would give $x_0=\sqrt{101}$ which could be a good estimate for Newton method which would give as first iterate $$x_1=\frac{2 x_0-1}{2 \log (x_0)+2-\log (101)}=\sqrt{101}-\frac 12\approx 9.54988$$ Just added for your curiosity, still using the truncated expansion, the solution of the equation $2 x \log (x)-x \log (101)+1=0$ is given in terms of Lambert function $$x=-\frac{1}{2 W\left(-\frac{1}{2 \sqrt{101}}\right)}\approx 9.53654$$

Using the gamma function and calculus the exact solution would be $x=9.54573$