Use the integral test to determine whether the series converges.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 8 } { \sqrt { n } }$$
A) converges
B) diverges

Question

Quizplus · Accepted Answer

We can use the integral test to determine whether the series converges or diverges. Let $f(x) = \frac{8}{\sqrt{x}}$. Then $f(x)$ is a positive, continuous, and decreasing function for $x\geq 1$.
Using integration by substitution, we have:
$$\int_{1}^{n}\frac{8}{\sqrt{x}}dx = \left[16\sqrt{x}ight]_{1}^{n} = 16\sqrt{n} - 16$$
Since $f(x)$ is positive and decreasing, we have:
$$\int_{1}^{n}f(x)dx \geq \sum_{k=2}^{n}f(k) \geq \int_{2}^{n+1}f(x)dx$$
$$16\sqrt{n} - 16 \geq \sum_{k=2}^{n}\frac{8}{\sqrt{k}} \geq 16\sqrt{n+1} - 16\sqrt{2}$$
As $nightarrow\infty$, both sides approach infinity. Therefore, the series $\sum_{n=1}^{\infty}\frac{8}{\sqrt{n}}$ diverges by the integral test. Thus, the answer is $\boxed{	extbf{(B)}}$.

Use the Integral Test to Determine Whether the Series Converges ∑n=1∞8n\sum _ { n = 1 } ^ { \infty } \frac { 8 } { \sqrt { n } }n=1∑∞​n​8​

Use the Integral Test to Determine Whether the Series Converges $\sum _ { n = 1 } ^ { \infty } \frac { 8 } { \sqrt { n } }$