College Algebra 7th Edition

$\sum_{k=2}^{100}\frac{(-1)^{k}}{k\ln k}$
We write the sum in sigma notation: (We notice that the denominator takes the form $n\ln n$ with $n$ being an integer, while the numerator alternates between $1$ and $-1$.) $\frac{\mathrm{l}}{2\ln 2}-\frac{\mathrm{l}}{3\ln 3}+\frac{\mathrm{l}}{4\ln 4}-\frac{\mathrm{l}}{5\ln 5}+\cdots+\frac{\mathrm{l}}{100\ln 100}=\sum_{k=2}^{100}\frac{(-1)^{k}}{k\ln k}$