Show that the 3 n th convergent to 1 5- 1 2- 1 1- 1 5- 1 2- 1 1- 1 5- - is n 3n-1-

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Summation by Sigma Method

Show that the...

Question

Show that the $3 n^{t h}$ convergent to
$\frac{1}{5 -} \frac{1}{2 -} \frac{1}{1 -} \frac{1}{5 -} \frac{1}{2 -} \frac{1}{1 -} \frac{1}{5 -} \dots$ is $\frac{n}{3 n + 1}$ .

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Solution

We have
$p_{3 n} = p_{3 n - 1} - p_{3 n - 2};$
$p_{3 n - 1} = 2 p_{3 n - 2} - p_{3 n - 3};$
$p_{3 n - 2} = 5 p_{3 n - 3} - p_{3 n - 4};$
$p_{3 n - 3} = p_{3 n - 4} - p_{3 n - 5};$
$p_{3 n - 4} = 2 p_{3 n - 5} - p_{3 n - 6}$
From the first three equations, $p_{3 n} = 4 p_{3 n - 3} - p_{3 n - 4};$
From the last two equations,
$2 p_{3 n - 3} = p_{3 n - 4} - p_{3 n - 6}$
By combining these results we have,
$p_{3 n} = 2 p_{3 n - 3} - p_{3 n - 6}$
So that the scale of relation is $1 - 2 x + x^{2}$
$p_{3} = 1, q_{3} = 4, p_{6} = 2, q_{6} = 7.....$
$∴ p_{3} + p_{6} x + p_{9} x^{2} + . . . . . + p_{3 n} x^{n - 1} + . . . . = \frac{p_{3} + (p_{6} - 2 p_{3}) x}{1 - 2 x + x^{2}} = \frac{1}{1 - 2 x + x^{2}}$
Similarly, $q_{3} + q_{6} x + . . . . + q_{3 n} x^{n - 1} + . . . . .$
$= \frac{4 - x}{1 - 2 x + x^{2}}$
$∴ p_{3 n} = n,$
$q_{3 n} = 4 n - (n - 1) = 3 n + 1$

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Show that the $3 n^{t h}$ convergent to
$\frac{1}{5 -} \frac{1}{2 -} \frac{1}{1 -} \frac{1}{5 -} \frac{1}{2 -} \frac{1}{1 -} \frac{1}{5 -} \dots$ is $\frac{n}{3 n + 1}$ .