Sequence (01)

Real Analysis
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Grigorios Kostakos
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Sequence (01)

#1

Post by Grigorios Kostakos »

Let \(\{{a_{n}}\}_{n=1}^{\infty}\) a sequence of real numbers such that
\[0<a_{1}<a_2\quad {\text{ and}}\quad a_{n+1}=\sqrt{a_{n}\,a_{n-1}}\, , \;n\geqslant2\, .\]
a) Prove that the sequence \(\{{a_{n}}\}_{n=1}^{\infty}\) converges.
b) Prove that \(\displaystyle\mathop{\lim}\limits_{n\rightarrow{+\infty}}{a_{n}}=\sqrt[3]{a_2^2\,a_1}\) .
c) Find a closed form for \(a_{n}\) .
Grigorios Kostakos
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Grigorios Kostakos
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Re: Sequence (01)

#2

Post by Grigorios Kostakos »

c) For the sequence \(\{{a_{n}}\}_{n=1}^{\infty}\) of real numbers holds
$$0<a_{1}<a_2\quad {\text{ and}}\quad a_{n+1}=\sqrt{a_{n}\,a_{n-1}}\, , \quad(1) \quad n\geqslant2\, .$$
From \((1)\) we have
\begin{alignat*}{2}
\log a_{n+1}& =\log\sqrt{a_{n}\,a_{n-1}}=\log\bigl({\sqrt{a_{n}}\,\sqrt{a_{n-1}}\,}\bigr)=\log\bigl({a_{n}^{\frac{1}{2}}\,a_{n-1}^{\frac{1}{2}}}\bigr)\\
& =\log a_{n}^{\frac{1}{2}}+\log a_{n-1}^{\frac{1}{2}}=\frac{1}{2}\,\log a_{n}+\frac{1}{2}\,\log a_{n-1}\,.
\end{alignat*}
Substituting \(x_{n}=\log a_{n}\,,\) we have
$$x_{n+1}=\displaystyle\frac{1}{2}\,x_{n}+\frac{1}{2}\,x_{n-1}\,.$$
It is known that
$$x_{n}=\displaystyle\frac{2}{3}\,\Bigl({1-\Bigl({-\frac{1}{2}}\Bigr)^{n-1}}\Bigr)\,x_2+\frac{2}{3}\,\Bigl({\frac{1}{2}+\Bigl({-\frac{1}{2}}\Bigr)^{n-1}}\Bigr)\,x_{1}\,.$$
So
\begin{alignat*}{2}
\log a_{n} & =\displaystyle\frac{2}{3}\,\Bigl({1-\Bigl({-\frac{1}{2}}\Bigr)^{n-1}}\Bigr)\,\log a_{2}+\frac{2}{3}\,\Bigl({\frac{1}{2}+\Bigl({-\frac{1}{2}}\Bigr)^{n-1}}\Bigr)\,\log a_{1}& \\
& =\log a_{2}^{\frac{2}{3}\bigl({1-({-\frac{1}{2}})^{n-1}}\bigr)}+\log a_{1}^{\frac{2}{3}\bigl({\frac{1}{2}+({-\frac{1}{2}})^{n-1}}\bigr)}& \\
& =\log\Bigl({a_{2}^{\frac{2}{3}\bigl({1-({-\frac{1}{2}})^{n-1}}\bigr)}\,a_{1}^{\frac{2}{3}\bigl({\frac{1}{2}+({-\frac{1}{2}})^{n-1}}\bigr)}}\Bigr)\quad\Rightarrow & \\
a_{n} &=a_{2}^{\frac{2}{3}\bigl({1-({-\frac{1}{2}})^{n-1}}\bigr)}\,a_{1}^{\frac{2}{3}\bigl({\frac{1}{2}+({-\frac{1}{2}})^{n-1}}\bigr)}\,,\quad n\geqslant3\,. &\hspace{2.8cm}\square
\end{alignat*}
Grigorios Kostakos
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