Answer
$\ln 2-\ln 3+\ln 4-...+(-1)^{n}\ \ln n$
Work Step by Step
We see that there are $n-1$ terms, as the index $k$ changes from $2$ to $n$. The index $k$ indicates how the terms are formed.
Thus, we write the sum for the $n$ terms as follows:
$\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) $
We find that we have alternating signs. The value of $(-1)^{n}$ is $-1$ when $n$ is odd and $+1$ when $n$ is even. Thus, we write:
$=\ln 2-\ln 3+\ln 4-...+(-1)^{n}\ \ln n$
