Let $n\in \mathbb{Z}_{+}^{*}$ and let ${{\left( {{f}_{n}} \right)}_{n\ge 0}}$ be a sequence of functions defined over $\left( 0,1 \right)$ in such way ${{f}_{n}}\left( t \right)=\frac{1}{2t}\int_{0}^{1}{\left| {{x}^{n}}-t \right|dx}+\frac{1}{2}$ with ${{a}_{n}}$ is the minimum of ${{f}_{n}}$
Show that, $\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{n\le i\le 2n-1}{{{a}_{i}}}=\frac{1}{{{2}^{\ln 2}}}$
Solution: from the definition of absolute value we have
$\left| {{x}^{n}}-t \right|=\left\{ \begin{align}
& {{x}^{n}}-t\,\,,\,\,\,{{t}^{1/n}}\le x<1 \\
& 0\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,x={{t}^{1/n}} \\
& t-{{x}^{n}}\,\,\,,\,\,\,\,0<x<{{t}^{1/n}} \\
\end{align} \right.$
Hence,${{f}_{n}}\left( t \right)=\frac{1}{2t}\int_{0}^{{{t}^{1/n}}}{\left( t-{{x}^{n}} \right)dx}+\frac{1}{2t}\int_{{{t}^{1/n}}}^{1}{\left( {{x}^{n}}-t \right)dx}+\frac{1}{2}$
$=\frac{1}{2t}\left[ \left( tx \right)_{0}^{{{t}^{1/n}}}-\frac{1}{n+1}\left( {{x}^{n+1}} \right)_{0}^{{{t}^{1/n}}} \right]+\frac{1}{2t}\left[ \frac{1}{n+1}\left( {{x}^{n+1}} \right)_{{{t}^{1/n}}}^{1}-\left( tx \right)_{{{t}^{1/n}}}^{1} \right]+\frac{1}{2}$
$=\frac{1}{2t}\left[ {{t}^{1/n+1}}-\frac{1}{n+1}{{t}^{\frac{n+1}{n}}} \right]+\frac{1}{2t}\left[ \frac{1}{n+1}\left( 1-{{t}^{\frac{n+1}{n}}} \right)-\left( t-{{t}^{1/n+1}} \right) \right]+\frac{1}{2}$
$=\frac{1}{2t}\left[ {{t}^{1/n+1}}-\frac{{{t}^{\frac{n+1}{n}}}}{n+1}+\frac{1}{n+1}-\frac{{{t}^{\frac{n+1}{n}}}}{n+1}-t+{{t}^{1/n+1}} \right]+\frac{1}{2}$
$=\frac{1}{2t}\left[ 2t.{{t}^{1/n}}-\frac{2}{n+1}{{\left( {{t}^{n+1}} \right)}^{1/n}}+\frac{1}{n+1}-t \right]+\frac{1}{2}={{t}^{1/n}}-\frac{1}{\left( n+1 \right)}\left( {{t}^{1/n}} \right)+\frac{1}{2t\left( n+1 \right)}-\frac{1}{2}+\frac{1}{2}$
$={{t}^{1/n}}\left( 1-\frac{1}{n+1} \right)+\frac{1}{2t\left( n+1 \right)}=\frac{n}{n+1}{{t}^{1/n}}+\frac{1}{2t\left( n+1 \right)}=\frac{1}{n+1}\left[ n{{t}^{1/n}}+\frac{1}{2t} \right]$
Hence ${{f}_{n}}\left( t \right)=\frac{1}{n+1}\left( n{{t}^{1/n}}+1/2t \right)$ but ${{a}_{n}}$ is the minimum of ${{f}_{n}}$
Hence ${{f}_{n}}'\left( t \right)=\frac{1}{n+1}\left( {{t}^{1/n-1}}-\frac{1}{2{{t}^{2}}} \right)\Leftrightarrow {{f}_{n}}'\left( t \right)=0\Leftrightarrow {{t}^{1/n-1}}=\frac{1}{2{{t}^{2}}}\Leftrightarrow t={{\left( \frac{1}{2} \right)}^{1/n+1}}$
So that ${{f}_{n}}$ attend a local max or min at $t={{\left( 1/2 \right)}^{1/n+1}}$
But ${{f}_{n}}''\left( t \right)=\frac{1}{n+1}\left( \left( 1/n-1 \right){{t}^{1/n-2}}+\frac{1}{{{t}^{3}}} \right)$
We have $0<t<1\Leftrightarrow 0<{{t}^{1/n-2}}<1\And 0<{{t}^{3}}<1\Leftrightarrow {{t}^{-3}}>1$ hence ${{f}_{n}}''\left( t \right)>0$
This proves that ${{f}_{n}}$ has local minimum \[{{a}_{n}}=\underset{0<t<1}{\mathop{\min }}\,{{f}_{n}}\left( t \right)={{f}_{n}}\left( {{\left( 1/2 \right)}^{1/n+1}} \right)={{\left( 1/2 \right)}^{\frac{1}{n+1}}}\]
Let $L=\underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{n\le i\le 2n-1}{{{a}_{i}}\Leftrightarrow \ln L=\ln }\left( \underset{n\to \infty }{\mathop{\lim }}\,\prod\limits_{n\le i\le 2n-1}{{{a}_{i}}} \right)=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=n}^{2n-1}{\ln {{a}_{i}}}$
Notice that, $\ln {{a}_{n}}=\frac{1}{n+1}\ln \left( 1/2 \right)=-\frac{\ln 2}{n+1}$ hence
$\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=n}^{2n-1}{\ln {{a}_{i}}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=n}^{2n-1}{\frac{\ln {{2}^{-1}}}{i}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{i=1}^{n}{\frac{\ln {{2}^{-1}}}{n+i}}=-\ln 2\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=1}^{n}{\frac{1}{n+k}}$
Notice that, $A=\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=1}^{n}{\frac{1}{n+k}}$ is Riemann sum with $\Delta {{x}_{i}}=\frac{k}{n}$ and $h=\frac{1-0}{n}$
In partition $0={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<....<{{x}_{n}}=1$ and ${{x}_{i}}=a+ih$
Hence $\underset{n\to \infty }{\mathop{\lim }}\,\sum\nolimits_{k=1}^{n}{\frac{1}{n+k}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\nolimits_{k=1}^{n}{\frac{1}{1+k/n}=\int_{0}^{1}{\frac{1}{x+1}dx}=\ln \left| x+1 \right|_{0}^{1}=\ln 2}}$
Thus $\ln L=-\ln 2\ln 2=\ln {{2}^{-\ln 2}}\Leftrightarrow L=\frac{1}{{{2}^{\ln 2}}}$
Idea of solution Credit to Dr. B. Barakat
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