# RMO Training: taking help from Nordic mathematical contest: 1988

Problem:

Let $m_{n}$ be a smallest value of the function $f_{n}(x)=\sum_{k=0}^{2n}x^{k}$. Prove that $m_{n} \rightarrow \frac{1}{2}$ when $n \rightarrow \infty$.

Proof:

For $n>1$,

$f_{n}(x)=1+x+x^{2}+\ldots=1+x(1+x+x^{2}+x^{4}+\ldots)+x^{2}(1+x^{2}+x^{4}+\ldots)=1+x(1+x)\sum_{k=0}^{n-1}x^{2k}$.

From this, we see that $f_{n}(x)\geq 1$ for $x \leq -1$ and $x\geq 0$. Consequently, $f_{n}$ attains its maximum value in the interval $(-1,0)$. On this interval

$f_{n}(x)=\frac{1-x^{2n+1}}{1-x}>\frac{1}{1-x}>\frac{1}{2}$

So, $m_{n} \geq \frac{1}{2}$. But,

$m_{n} \leq f_{n}(-1+\frac{1}{\sqrt{n}})=\frac{1}{2-\frac{1}{\sqrt{n}}}+\frac{(1-\frac{1}{\sqrt{n}})^{2n+1}}{2-\frac{1}{\sqrt{n}}}$

As $n \rightarrow \infty$, the first term on the right hand side tends to the limit $\frac{1}{2}$. In the second term, the factor

$(1-\frac{1}{\sqrt{n}})^{2n}=((1-\frac{1}{\sqrt{n}})^{\sqrt{n}})^{2\sqrt{n}}$

of the numerator tends to zero because

$\lim_{k \rightarrow \infty}(1-\frac{1}{k})^{k}=e^{-1}<1$.

So, $\lim_{n \rightarrow \infty}m_{n}=\frac{1}{2}$

auf wiedersehen,

Nalin Pithwa.

Reference: Nordic Mathematical Contest, 1987-2009.

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