How do you use the integral test to determine the convergence or divergence of #1+1/root3(4)+1/root3(9)+1/root3(16)+1/root3(25)+...#?

Question

Samantha Adkison · Accepted Answer

The series:

#sum_(n=1)^oo 1/root(3)(n^2)#

is divergent.

We can write the series synthetically as: #sum_(n=1)^oo 1/root(3)(n^2) = sum_(n=1)^oo 1/n^(2/3)# This is series is divergent based on the p-series test as #2/3 < 1#

Ezra Adams · Answer

undefined To use the integral test to determine the convergence or divergence of the series $1 + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{9}} + \frac{1}{\sqrt{16}} + \frac{1}{\sqrt{25}} + \dots$, follow these steps:

1. Define the function $f(x)$ to be the terms of the series: $f(x) = \frac{1}{\sqrt{x}}$.

2. Verify that the function $f(x)$ is continuous, positive, and decreasing for $x \geq 1$, which is the domain relevant to our series.

3. Apply the integral test:
   - Evaluate the improper integral $ \int_{1}^{\infty} f(x) \, dx = \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx$.

4. Integrate $ \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx$:
   $$
   \int_{1}^{\infty} \frac{1}{\sqrt{x}} \, dx = \lim_{b \to \infty} \left[2\sqrt{x}\right]_{1}^{b} = \lim_{b \to \infty} \left(2\sqrt{b} - 2\right)
   $$

5. Determine the convergence of the integral:
   - Since the limit exists and is finite, the integral converges.

6. Conclusion:
   - By the integral test, since the integral converges, the original series $1 + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{9}} + \frac{1}{\sqrt{16}} + \frac{1}{\sqrt{25}} + \dots$ also converges.