Question 1069243
It is an infinite geometric series with common ratio
r = {{{ln(x)/2}}}
The series converges if
|r| < 1
{{{ln(x)/2}}} < 1
ln(x) < 2
x < {{{e^2}}}
Sum of series = {{{a/(1-r)}}}
= {{{1/(1-ln(x)/2)}}}
When x = {{{e^(1/2)}}}, then the sum of the series
= {{{1/(1-1/4)}}}
= {{{1/(3/4)}}}
= {{{4/3}}}