Answer
$L \approx 28.369$
Work Step by Step
We have from Exercise 90:
$G\left( k \right) = \dfrac{\pi }{2} - \dfrac{\pi }{2}\mathop \sum \limits_{n = 1}^\infty {\left( {\dfrac{{1\cdot3\cdot\cdot\cdot\left( {2n - 1} \right)}}{{2\cdot\cdot\cdot4\cdot\left( {2n} \right)}}} \right)^2}\dfrac{{{k^{2n}}}}{{2n - 1}}$, ${\ \ \ }$ converges for $\left| k \right| \lt 1$.
We use the first three terms of the expansion, so
$G\left( k \right) \approx \dfrac{\pi }{2} - \dfrac{\pi }{2}\left[ {{{\left( {\dfrac{1}{2}} \right)}^2}{k^2} + {{\left( {\dfrac{{1\cdot 3}}{{2\cdot 4}}} \right)}^2}\dfrac{{{k^4}}}{3} + {{\left( {\dfrac{{1\cdot 3\cdot 5}}{{2\cdot 4\cdot 6}}} \right)}^2}\dfrac{{{k^6}}}{5}} \right]$
It is known that the arc length of the ellipse is $L = 4bG\left( k \right)$, where $k = \sqrt {1 - \dfrac{{{a^2}}}{{{b^2}}}} $.
When $a=4$ and $b=5$, we get $k = \dfrac{3}{5}$. Thus,
$L = 20G\left( {\dfrac{3}{5}} \right) \approx \dfrac{\pi }{2} - \dfrac{\pi }{2}\left[ {{{\left( {\dfrac{1}{2}} \right)}^2}{{\left( {\dfrac{3}{5}} \right)}^2} + {{\left( {\dfrac{{1\cdot3}}{{2\cdot4}}} \right)}^2}\dfrac{{{{\left( {\dfrac{3}{5}} \right)}^4}}}{3} + {{\left( {\dfrac{{1\cdot3\cdot5}}{{2\cdot4\cdot6}}} \right)}^2}\dfrac{{{{\left( {\dfrac{3}{5}} \right)}^6}}}{5}} \right]$
We found that $L \approx 28.369$.
