The series is $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}k}{k^2+1}$
The minimum number of terms needed should be given by $\lvert a_{n+1} \rvert$
Thus $\lvert \frac{(-1)^{k+2}(k+1)}{(k+1)^2+1} \rvert<0.01$
However, this is only satisfied after the $99th$ term, whereas the answer says the $19th$ term.
For 1dp the bound should be.05, not .01.