Answer
$$\dfrac{61}{36}$$
Work Step by Step
We simplify the summation by plugging in the values from $1$ to $3$ as follows:
$$\displaystyle \sum_{k=1}^{3} (-1)^{k+1}\frac{k+1}{k^2}\\=(-1)^{1+1}\ \frac{1+1}{1^2}+(-1)^{2+1}\ \frac{2+1}{2^2}+(-1)^{3+1}\ \frac{3+1}{3^2}\\=(-1)^{2}\times \frac{2}{1}+(-1)^{3}\times \frac{3}{4}+(-1)^{4}\times \frac{4}{9}\\=\frac{61}{36}$$
