Question

$\ T_{k} = ( - 1)^{k + 1} \cdot \frac{k(k + 1)}{\lfloor k} = ( - 1)^{k + 1}\left( \frac{k(k - 1) + 2k}{\lfloor k} \right)$

∴ sum $= \sum_{k = 1}^{\infty}\mspace{2mu}\frac{( - 1)^{k + 1}}{k - 2} + \sum_{k = 1}^{\infty}\mspace{2mu}\frac{2( - 1)^{k + 1}}{k - 1}$

$${= \left( \frac{1}{⌞ - 1} - \frac{1}{⌞0} + \frac{1}{⌞1} - \frac{1}{⌞2} + \frac{1}{⌞3}\ldots \right) + \left( \frac{2}{⌞0} - \frac{2}{⌞1} + \frac{2}{⌞2} - \frac{2}{⌞3}\ldots \right) }{= \frac{1}{e}}$$