Q.2.b) - Calculus

Question

As a p- series you could use a comparison to show that the series $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k}\right)$ diverges.

Step-by-Step Solution

Answer

The $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k}\right)$is divergent

The objective is to find the p- series that is used to show that the series ∑ k = 1 ∞ sin 1 k is convergent

The comparison test is used to determine the convergence or divergence of the series

It states that $\sum _{k=1}^{\infty }{a}_k$and $\sum _{k=1}^{\infty }{b}_k$ be two series with positive terms such that $0\le a_k\le b_k$ for every positive integer k.

If the series $\sum _{k=1}^{\infty }{b}_k$converges then the series $\sum _{k=1}^{\infty }{a}_k$also convergences

According to the given data

The term of series $\underset{k=1}{\overset{\infty }{\sum \sin \left(\frac{1}{k}\right)}}$ are positive

The expression of $\sin \left(\frac{1}{k}\right)$ follows inequality

$\sin \left(\frac{1}{k}\right)\le \frac{1}{k}$

The series $\sum _{k=1}^{\infty }{b}_k$for the series $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k}\right)$ is given by

$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$

By concluding

The series $\sum _{k=1}^{\infty }{b}_k$=$\sum _{k=1}^{\infty }\frac{1}{k}$ is divergent by the p- series

Therefore the$\sum _{k=1}^{\infty }{a}_k$ is  also divergent

Hence fore the $\sum _{k=1}^{\infty }\sin \left(\frac{1}{k}\right)$ is divergent and p-series is$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$