# What is the maximum value of $\frac{x^{100}}{1+x+x^2+\ldots+x^{200}}$?

If $$x$$ is positive, what is the maximum value of this expression:

$$\frac{x^{100}}{1+x+x^2+\ldots+x^{200}}$$

This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.

This is what I have tried:

The denominator is a geometric series whose sum is

$$\frac{1-x^{201}}{1-x}$$

The fraction now becomes

$$\frac{x^{100}(1-x)}{1-x^{201}}$$

I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $$x$$ and applying the AM-GM-HM inequality.

That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).

I can't see such means from looking at the fraction. Can someone help me here?

• For $x>0$ we have$$\sum_{n=0}^{200}x^n\ge 201 \sqrt[201]{\prod_{n=0}^{200}x^n}=201 \sqrt[201]{x^{20100}}=201x^{100}$$ $$\frac{x^{100}}{\sum_{n=0}^{200}x^n}\le \frac1{201}$$ – Mark Viola Jan 30 at 4:45

The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $$201$$ terms is $$201$$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.

You can instead minimise the reciprocal of your quantity, viz., $$\frac{1+x+x^2+\cdots+x^{200}}{x^{100}}=x^{-100}+x^{-99}+\cdots+x^{99}+x^{100}.$$ One only needs the two-variable AM/GM inequality to do this, just in the form $$y+y^{-1}\ge2$$ for $$y>0$$, for $$x^{-100}+x^{-99}+\cdots+x^{99}+x^{100}=1+\sum_{n=1}^{100}(x^n+x^{-n}) \ge201$$ with equality if $$x=1$$.

For $$x>0$$, we have from the AM-GM inequality

\begin{align} \sum_{n=0}^{200}x^n&\ge 201 \sqrt[201]{\prod_{n=0}^{200}x^n}\\\\ &=201 \sqrt[201]{x^{20100}}\\\\ &=201x^{100} \end{align}

Hence, we see that

$$\frac{x^{100}}{\sum_{n=0}^{200}x^n}\le \frac1{201}$$

• Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Jan 30 at 5:02