Question

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

Answer 1

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