Question

please show that the sequence a(n+1)=sqr(3an) where a1=2 is convergent and find its limit.

Answer 1

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Answer 2

To show the sequence converges, you have to establish monotonicity and boundedness.

To see if there's an initial pattern of monotonicity, compare the first two terms:
\[a_2=\sqrt{3a_1}=\sqrt{3\times2}=\sqrt6>\sqrt4=2=a_1\]
So we know the sequence is increasing at least between the first two terms. This is the basis case (\(n=1\)).

Assume \(a_{k+1}>a_k\), and show that \(a_{k+2}>a_{k+1}\).
\[\begin{align*}
a_{k+2}&=\sqrt{3a_{k+1}}\\
&>\sqrt{3a_k}\\
&=a_{k+1}
\end{align*}\]
and so the sequence is increasing for all \(n\ge1\).

To establish boundedness, you can notice as sort of pattern:
\[\begin{array}{cr}a_1&&2<3\\
a_2&&\sqrt6<\sqrt9=3\\
a_3&&\sqrt{3\sqrt6}=\sqrt[4]{54}<\sqrt[4]{81}=3\\
a_4&&\sqrt{3\sqrt[4]{54}}=\sqrt[8]{4374}<\sqrt[8]{6561}=3
\end{array}\]
and so on. So we know that 3 is an upper bound.

The sequence is bounded and monotonic, so it must converge.