Chapter 8.3, Problem 12E

To determine

To find: whether the series is convergent or divergent.

Answer to Problem 12E

The series is convergent

Explanation of Solution

Given information: The series is ${\displaystyle \sum _{n=1}^{\infty }(n^{-1.4}}-3n^{-1.2})$

Given, ${\displaystyle \sum _{n=1}^{\infty }(n^{-1.4}}-3n^{-1.2})$

Let, $u_n={\displaystyle \sum _{n=1}^{\infty }(n^{-1.4}}-3n^{-1.2})$

From the definition of partial sum , we have $u_n={\displaystyle \sum _{n=1}^{\infty }{n}^{-1.4}}-{\displaystyle \sum _{n=1}^{\infty }3n^{-1.2}}$

Also let, $a_n=n^{-1.4}$ and $b_n=3n^{-1.2}$

Now we find convergence or divergence of $a_n$

  $a_n=n^{-1.4}$

  $\Rightarrow a_n=\left(\frac{1}{n^{{}^{1.4}}}\right)$

By the theorem of p series we can say that the series is convergent where $1.4(p\text{ say)}\ge 1$ .

Similarly we say that the series $b_n=3n^{-1.2}$ is convergent .

Sum of two convergent series is convergent.

Hence , the series is convergent.