Chapter 12, Problem 3CT

To determine

To find: The sum of the sequence.

Answer to Problem 3CT

The sum of the sequence is ${{\displaystyle \sum _{k=1}^3\left(-1\right)}}^{k+1}\left(\frac{k+1}{k^2}\right)=\frac{61}{36}$

Explanation of Solution

Given information:

The given Sum is,

  ${{\displaystyle \sum _{k=1}^3\left(-1\right)}}^{k+1}\left(\frac{k+1}{k^2}\right)$

Calculation:

Substitute $1,2,$ and $3$ for $k$ in the sequence ${{\displaystyle \sum _{k=1}^3\left(-1\right)}}^{k+1}\left(\frac{k+1}{k^2}\right)$ .

  $\begin{array}{c}{{\displaystyle \sum _{k=1}^3\left(-1\right)}}^{k+1}\left(\frac{k+1}{k^2}\right)={\left(-1\right)}^{1+1}\cdot \left(\frac{1+1}{1^2}\right)+{\left(-1\right)}^{2+1}\cdot \left(\frac{2+1}{2^2}\right)+{\left(-1\right)}^{3+1}\cdot \left(\frac{3+1}{3^2}\right)\\ ={\left(-1\right)}^2\cdot \frac{2}{1}+{\left(-1\right)}^3\cdot \left(\frac{3}{4}\right)+{\left(-1\right)}^4\cdot \left(\frac{4}{9}\right)\\ =1\cdot \left(2\right)+\left(-1\right)\cdot \frac{3}{4}+1\cdot \left(\frac{4}{9}\right)\\ =\frac{61}{36}\end{array}$

Therefore, the evaluate sum is ${{\displaystyle \sum _{k=1}^3\left(-1\right)}}^{k+1}\left(\frac{k+1}{k^2}\right)=\frac{61}{36}$ .