sequence

[logistic_guy]

Consider the sequence defined recursively by \(\displaystyle x_n = \sqrt{\sqrt{1996 x_{n-1}}}\) and \(\displaystyle x_0 = 1\). Calculate the first few terms of this sequence, and decide whether it approaches a limiting value.

 

[khansaheb]

Consider the sequence defined recursively by \(\displaystyle x_n = \sqrt{\sqrt{1996 x_{n-1}}}\) and \(\displaystyle x_0 = 1\). Calculate the first few terms of this sequence, and decide whether it approaches a limiting value.

show us your effort/s to solve this problem.

 

[logistic_guy]

\(\displaystyle x_1 \approx \sqrt{\sqrt{1996}} = 6.68406\)

\(\displaystyle x_2 \approx \sqrt{\sqrt{1996 \times 6.68406}} = 10.74732\)

\(\displaystyle x_3 \approx \sqrt{\sqrt{1996 \times 10.74732}} = 12.10222\)

\(\displaystyle x_4 \approx \sqrt{\sqrt{1996 \times 12.10222}} = 12.46684\)

The sequence suddenly started to increase slowly which suggests that there might be a limit value.
Let us assume that there is a limit \(\displaystyle x_n \rightarrow L\) as \(\displaystyle n \rightarrow \infty\).

Then,

\(\displaystyle L = \sqrt{\sqrt{1996L}}\)

\(\displaystyle L^2 = \sqrt{1996L}\)

\(\displaystyle L^4 = 1996L\)

\(\displaystyle L^3 = 1996\)

This means that the limit of the sequence is:

\(\displaystyle L = \sqrt[3]{1996} \approx 12.59081\)