Using the integral test, how do you show whether ##sum 1/sqrt(n+1)## diverges or converges from n=1 to infinity?

The integral test basically works from the definition of the integral (quick version: the integral is the accumulated sum of infinitely thin differential intervals ##dn## over a specified interval ##a->b##).

A paraphrased version of the integral test is as follows:

Let there be a function ##f(n) = a_n## where ##a_n## is a series lying within the domain ##[k,oo)##. There exists another function ##f(x)## that is continuous, positive, and decreasing such that the convergence or divergence of ##int_k^(oo)f(x)dx## determines the convergence or divergence of ##sum_(n=k)^(oo)a_n##, respectively.

So, essentially, we have to integrate this, which is indeed continuous, positive, and decreasing at ##[1,oo)##:

##int_1^(oo)1/(sqrt(x+1))dx##

We can do that like so:

##= int_1^(oo)(x+1)^(-"1/2")dx##

##= |[2(x+1)^("1/2")]|_1^(oo)##

At this point we know that ##sqrt(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