1428. Given a convergent alternating series $\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n^{2}}$. Estimate the error made when replacing the sum of this series: 1) by the sum of its first three terms; 2) by the sum of its first four terms.
Solution. 1) $S=S_{n}+r_{n}$, where $S$ is the sum of the series; $S_{n}$ is the partial sum; $r_{n}$ is the remainder of the series. We find $S=S_{3}+r_{3}$ or $S=1-\frac{1}{2^{2}}+\frac{1}{3^{2}}+r_{3}$, from which $S=\frac{31}{36}+r_{3}$. By the Leibniz theorem, the absolute error $\left|r_{n}\right|<\frac{1}{(n+1)^{2}}$ (for $n>0$), therefore; $0<r_{4}<0.04$. Thus, the sum of the series $\mathrm{S} \approx \frac{115}{144}$ with a deficit.