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.

Short answer

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

Step-by-step solution

The objective is to find the p- series that is used to show that the series ∑k=1∞sin1k 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$\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k}$