Lance S. answered 04/10/20
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your using the right parent function to compare it to 1/(x^2). but as x goes to infinity then the entire function will tend to 0. The comparison test tells u to look at ur original function 1/(1+x^2) and determine if it is larger than the parent function or less than the parent function. in our case
1/(1+x^2) < 1/(x^2) so the if 1/(x^2).converges then so must the baby function 1/(1+x^2) converge.
if the baby function were larger than the parent function say
1/x > 1/(x^2) then the comparison test would be invalid, which is not the case here

Lance S.
04/10/20
Christopher T.
Thanks!04/10/20
Christopher T.
Sorry, but I’m confused, isn’t the parent function an improper integral as x approaches 0, which makes the entire expression diverge not converge, sorry if I’m not understanding04/10/20