Question

Find an example of a continuous function f :$[1,\infty )\to R$such that ${\int }_1^{\infty }f\left(x\right)dx$diverges and localid="1649077247585" $\sum _{k=1}^{\infty }f\left(k\right)$converges.

Short answer

An example is$\sin \left(\mathrm{\pi x}\right)$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

The function is $[1,\infty )\to R$.

The given integral is${\int }_1^{\infty }f\left(x\right)dx$.

Step 2. Evaluate the integral.

Evaluating the integral,

$$\begin{gathered} {\int }_1^{\infty }f\left(x\right)dx={\int }_1^{\infty }\sin \left(\mathrm{\pi x}\right)dx \\ =\underset{k\to \infty }{\lim }{\int }_1^k\sin \left(\mathrm{\pi x}\right)dx \\ =\underset{k\to \infty }{\lim }{\int }_1^k{{\left[\frac{-\cos \left(\mathrm{\pi x}\right)}{\mathrm{\pi }}\right]}^k}_1 \\ =\underset{k\to \infty }{\lim }{\int }_1^k\left[\frac{-\cos \left(\mathrm{\pi x}\right)+\cos \pi }{\mathrm{\pi }}\right] \\ =\underset{k\to \infty }{\lim }{\int }_1^k\left[\frac{-\cos \left(\mathrm{\pi x}\right)-1}{\mathrm{\pi }}\right] \\ =\infty \end{gathered}$$

Thus, the improper integral ${\int }_1^{\infty }\sin \left(\mathrm{\pi x}\right)dx$is divergent.

Step 3. To prove the series as convergent.

The value of $\sum _{k=1}^{\infty }f\left(k\right)=\sum _{k=1}^{\infty }\sin \left(\pi k\right)$is given below,

$$\begin{gathered} \sum _{k=1}^{\infty }\sin \mathrm{\pi k}=\sin \mathrm{\pi }+\sin 2\mathrm{\pi }+\sin 3\mathrm{\pi }+... \\ \sum _{\mathrm{k}=1}^{\infty }\sin \mathrm{\pi k}=0+0+0+... \\ \sum _{\mathrm{k}=1}^{\infty }\sin \mathrm{\pi k}==0 \end{gathered}$$

Thus, the series $\sum _{k=1}^{\infty }f\left(k\right)=\sum _{k=1}^{\infty }\sin \left(\pi k\right)$is convergent.