Answer
$\ln 2-\ln 3+\ln 4-\cdots+(-1)^{n}\ln n$
Work Step by Step
There are n-1 terms, as the index k changes from 2 to n.
The index k dictates how the terms are formed:
$\displaystyle \sum_{k=2}^{n}(-1)^{k}\ln k=(-1)^{2}\ln 2+(-1)^{3}\ln 3+(-1)^{4}\ln 4+...(-1)^{n}\ln n$
When the exponent of $(-1)^{n}$ is odd, we have $-1$
and when it is even, $+1.$
$=\ln 2-\ln 3+\ln 4-\cdots+(-1)^{n}\ln n$
