Using the integral test, how do you show whether #sum 1/sqrt(2x-5)# diverges or converges from n=1 to infinity?

Question

Using the integral test, how do you show whether #sum  1/sqrt(2x-5)# diverges or converges from n=1 to infinity?

Isaiah Allen · Accepted Answer

The series #sum_{n=3}^{infty}1/sqrt{2n-5}# diverges by the integral test, therefore the series #sum_{n=1}^{infty}1/sqrt{2n-5}# diverges as well.

Let #f(x)=1/sqrt{2x-5}=(2x-5)^{-1/2}#. The improper integral #\int_{3}^{infty}f(x)\ dx# can be shown to diverge by direct calculation. First, do the indefinite integral #int (2x-5)^{-1/2}\ dx# by using the substitution: #u=2x-5#, #du=2\ dx# to write it as #1/2 int u^{-1/2}\ du=u^{1/2}+C=sqrt{2x-5}+C#. Now write the improper integral as a limit: #int_{3}^{infty}f(x)\ dx=lim_{b->infty}int_{3}^{b}(2x-5)^{-1/2}\ dx=lim_{b->infty} sqrt{2x-5}|_{x=3}^{x=b}# #=lim_{b->infty}(sqrt{2b-5}-1)# This last limit doesn't exist (some would say it "diverges to #infty#"). Since #f(x)>0# for all #x\geq 3# and since #f# is decreasing on the interval #x\geq 3#, it follows from the integral test that the series #sum_{n=3}^{infty}1/sqrt{2n-5}# diverges. Since the series in question #sum_{n=1}^{infty}1/sqrt{2n-5}# just contains two more terms at the beginning, it follows that #sum_{n=1}^{infty}1/sqrt{2n-5}# diverges as well. And actually, the first two terms of this series are complex numbers since #2n-5<0# when #n=1# and #n=2#. So maybe you meant something different than your original question anyway.

Samantha Adkison · Answer

undefined To determine whether the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{2n-5}}$ converges or diverges, we can use the integral test.

1. Let $f(x) = \frac{1}{\sqrt{2x-5}}$.
2. Check if $f(x)$ is continuous, positive, and decreasing for $x \geq 1$.
3. If $f(x)$ meets these conditions, then the series converges if and only if the integral $\int_{1}^{\infty} f(x) \, dx$ converges.

Let's check the conditions for $f(x)$:

- Continuity: $f(x) = \frac{1}{\sqrt{2x-5}}$ is continuous for $x \geq \frac{5}{2}$, which includes the interval $[1, \infty)$.
- Positivity: $f(x) > 0$ for $x > \frac{5}{2}$, so it's positive on $[1, \infty)$.
- Decreasing: To show $f(x)$ is decreasing, we can analyze its derivative.

$$
f'(x) = -\frac{1}{2\sqrt{(2x-5)^3}}
$$

$f'(x) < 0$ for $x > \frac{5}{2}$, indicating that $f(x)$ is decreasing on $[1, \infty)$.

Since $f(x)$ meets all the conditions, we can proceed with the integral test:

$$
\int_{1}^{\infty} \frac{1}{\sqrt{2x-5}} \, dx
$$

To evaluate this integral, make the substitution $u = 2x - 5$, which implies $du = 2dx$, so $dx = \frac{1}{2} du$.

The new integral becomes:

$$
\int_{-5}^{\infty} \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du
$$

This is a proper integral that can be evaluated:

$$
\lim_{t \to \infty} \int_{-5}^{t} \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du = \lim_{t \to \infty} \left[ \sqrt{u} \right]_{-5}^{t} = \lim_{t \to \infty} \left( \sqrt{t} - \sqrt{-5} \right) = \infty
$$

Since the integral diverges, by the integral test, the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{2n-5}}$ also diverges.