Chapter 12.1, Problem 59AYU

In Problems 51-60, write out each sum.

${\displaystyle \underset{k=2}{\overset{n}{\sum }}{\left(-1\right)}^k}\ln k$

To determine

To find: Each sum: ${\displaystyle \sum }_{k=2}^n{\left(-1\right)}^k\ln k$.

Answer to Problem 59AYU

${\displaystyle \sum }_{k=2}^n{\left(-1\right)}^k\ln k=\ln 2-\ln 3+\ln 4+\dots \dots +{\left(-1\right)}^n\ln n$

Explanation of Solution

Given:

${\displaystyle \sum }_{k=2}^n{\left(-1\right)}^k\ln k$

Calculation:

${\displaystyle \sum }_{k=2}^n{\left(-1\right)}^k\ln k={\left(-1\right)}^2\ln 2+{\left(-1\right)}^3\ln 3+{\left(-1\right)}^4\ln 4+\dots \dots +{\left(-1\right)}^n\ln n$

${\displaystyle \sum }_{k=2}^n{\left(-1\right)}^k\ln k=\ln 2-\ln 3+\ln 4+\dots \dots +{\left(-1\right)}^n\ln n$