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

1 Answer
Jun 10, 2015

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