00:01
In this question we will recall about the integral test.
00:06
Which states that if the an will be the pon still terms and then the an plus 1 will be smaller than the an, which is the decreasing sequence, and this will imply that the summation of the an will behave the same as the integral of the a.
00:31
This end will become to the dx now as n equal to k to infinity so this will be from k to infinity and now in this question we're given the summation of the n of n square plus 1 power 2 for n gets from 1 to infinity so we can identify the one will be the k and this one will be the a n so we see that all of the term will be positive and the next term will be smaller than the previous term.
01:06
So therefore the behavior of this series here will be the same as the integral from 1 to infinity.
01:13
And then we have n returns to the x over the x squared plus 1 square and then the x.
01:21
Now to do this integral we use the substitution where we put this one will be the ux.
01:28
So we'll be u equal to the x square plus 1, du equal to the 2x dx.
01:35
So we notice that the x dx, we exactly this one, so x dx we equal to the du over 2.
01:42
And as a result we can write this down to the integral of the du over 2 and we will have the u square.
01:53
And also we need to replace the limit as well...