Q. 7 - Calculus

Question

Let f(x) be a function that is continuous, positive, and decreasing on the interval $[1,\infty )$ such that $\underset{x\to \infty }{lim}f\left(x\right)=\alpha >0$, What can the integral tells us about the series $\sum _{k=1}^{\infty }f\left(k\right)$ ?

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\infty }f\left(k\right)$ is divergent.

Step 1. Given Information.

The function:

$\underset{x\to \infty }{lim}f\left(x\right)=\alpha >0 on [1,\infty )$

Step 2. Integral test.

By the integral test, the function will both converge or diverge.

$$\begin{gathered} {\int }_1^{\infty }f\left(x\right)dx=\underset{k\to \infty }{lim}{\int }_1^k\alpha .dx \\ =\underset{k\to \infty }{lim}{\left[\alpha x\right]}_1^k \\ =\underset{k\to \infty }{lim}[\alpha k-\alpha ] \\ =0 \end{gathered}$$

Step 3. Convergent or divergent.

By the integral test, the improper integral ${\int }_1^{\infty }f\left(x\right)dx$ is divergent.

So the series $\sum _{k=1}^{\infty }f\left(k\right)$ is divergent.