Answer
sum = $\displaystyle \sum_{k=1}^{100}\frac{(-1)^{k+1}}{k\ln k}$
Work Step by Step
The kth term $a_{k}$ has
$k\cdot\ln k$ in the denominator and
$\pm 1$ in the numerator, (odd k's are minus, even k's are plus),
so we write $(-1)^{k+1}$ for the numerator.
(when k is even, the exponent is odd, and when k is odd, the exponent is even)
$a_{k}=\displaystyle \frac{(-1)^{k+1}}{k\ln k}$
There are 100 terms (k=1,2,3,...,100).
sum = $\displaystyle \sum_{k=1}^{100}\frac{(-1)^{k+1}}{k\ln k}$
