The integral test basically works from the definition of the integral (quick version: the integral is the accumulated sum of infinitely thin differential intervals dndn over a specified interval a->ba→b).
A paraphrased version of the integral test is as follows:
Let there be a function f(n) = a_nf(n)=an where a_nan is a series lying within the domain [k,oo)[k,∞). There exists another function f(x)f(x) that is continuous, positive, and decreasing such that the convergence or divergence of int_k^(oo)f(x)dx∫∞kf(x)dx determines the convergence or divergence of sum_(n=k)^(oo)a_n∞∑n=kan, respectively.
So, essentially, we have to integrate this, which is indeed continuous, positive, and decreasing at [1,oo)[1,∞):
int_1^(oo)1/(sqrt(x+1))dx∫∞11√x+1dx
We can do that like so:
= int_1^(oo)(x+1)^(-"1/2")dx=∫∞1(x+1)−1/2dx
= |[2(x+1)^("1/2")]|_1^(oo)=∣∣[2(x+1)1/2]∣∣∞1
At this point we know that sqrt(x+1)√x+1 is a constantly increasing function, so it has an "open" accumulation that can never stop without a well-defined right-end boundary. Basically, it's a half-open integral that extends its domain forever and so it has no finite area.
= 2(oo)^("1/2") - cancel(2(1+1)^("1/2"))^"small"
=> oo
The integral does not converge, and so the series does not converge either. QED
