## Sunday, March 19, 2006

### To Infinity... And Beyond! Topic: Algebra/Inequalities. Level: Olympiad.

Problem: (1996 Ireland - #7) Prove that for all positive integers $n$,

$2^{\frac{1}{2}} \cdot 4^{\frac{1}{4}} \cdots (2^n)^{\frac{1}{2^n}} < 4$.

Solution: Clearly, the product increases, so if we show that the infinite product is equivalent to $4$, we know that the partial products must always be less than $4$. Rewrite the infinite product as

$2^{\frac{1}{2}} \cdot 2^{\frac{2}{4}} \cdot 2^{\frac{3}{8}} \cdots$.

We wish to show that

$\frac{1}{2}+\frac{2}{4}+\frac{3}{8} \cdots = 2$.

We can rewrite it as the sum of an infinite number of infinite geometric series, or

$\left(\frac{1}{2}+\frac{1}{4}+\cdots\right)+\left(\frac{1}{4}+\frac{1}{8}+\cdots\right)+\cdots$,

which becomes

$1+\frac{1}{2}+\frac{1}{4}+\cdots = 2$.

Hence the infinite product is equivalent to $2^2 = 4$ and the partial products are $< 4$ as desired. QED.

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Comment: With exponents, a lot of the time its a good idea to look at the exponents themselves because it usually reduces the problem to a sum instead of a product, which is easier to work with.

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Practice Problem: Evaluate $n^{\frac{1}{n}} \cdot (n^2)^{\frac{1}{n^2}} \cdot (n^3)^{\frac{1}{n^3}} \cdots$.