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Limit of the smallest function value

lfdahl

Well-known member
Nov 26, 2013
719
Let $m_n$ be the smallest value of the function:

$$f_n(x)=\sum_{k=0}^{2n}x^k.$$

Show, that $m_n\to\frac{1}{2}$ as $n \to \infty$.

Source: Nordic Math. Contest
 

lfdahl

Well-known member
Nov 26, 2013
719
Suggested solution:



For $n > 1$:

$$f_n(x) = 1 + x + x^2 + …$$
$$= 1+x(1 + x^2 +x^4 + …) + x^2(1 + x^2 +x^4 + ….)$$
\[= 1 + x(x+1)\sum_{k=0}^{n-1}x^{2k}\]

From this we see that $f_n(x) \geq 1$, for $x \leq −1$ and $x \geq 0$. Consequently, $f_n$ attains its minimum 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


\[m_n \leq f_n\left ( -1 + \frac{1}{\sqrt{n}}\right ) < \frac{1}{2-\frac{1}{\sqrt{n}}}+\frac{\left ( 1-\frac{1}{\sqrt{n}} \right )^{2n+1}}{2-\frac{1}{\sqrt{n}}}\]



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


\[\left ( 1-\frac{1}{\sqrt{n}} \right )^{2n} = \left ( \left ( 1-\frac{1}{\sqrt{n}} \right )^{\sqrt{n}} \right )^{2\sqrt{n}}\]

of the nominator tends to zero, because

\[\lim_{k\rightarrow \infty }\left ( 1-\frac{1}{k} \right )^k = e^{-1} < 1\]


Thus,

\[\lim_{n \rightarrow \infty }m_n = \frac{1}{2}.\]