问题及解答

用柯西审敛准则证明下列级数收敛.

1.    $\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}$

考虑 $S_{n+p}-S_n$,

\[
\begin{split}
S_{n+p}-S_n&=(-1)^{n}\frac{1}{n+1}+(-1)^{n+1}\frac{1}{n+2}+\cdots+(-1)^{n+p-1}\frac{1}{n+p}\\
&=(-1)^{n}\Bigl[\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\frac{1}{n+4}+\cdots+(-1)^{p-1}\frac{1}{n+p}\Bigr].
\end{split}
\]

因此, 当 $p=2k-1$ 时,

\[
\begin{split}
(-1)^n(S_{n+p}-S_n)&=\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\frac{1}{n+4}+\cdots+\frac{1}{n+2k-1}\\
&=\frac{1}{n+1}-(\frac{1}{n+2}-\frac{1}{n+3})-(\frac{1}{n+4}-\frac{1}{n+5})-\cdots-(\frac{1}{n+2k-2}-\frac{1}{n+2k-1})\\
&\leqslant\frac{1}{n+1},
\end{split}
\]

\[
\begin{split}
(-1)^n(S_{n+p}-S_n)&=(\frac{1}{n+1}-\frac{1}{n+2})+(\frac{1}{n+3}-\frac{1}{n+4})+\cdots+(\frac{1}{n+2k-3}-\frac{1}{n+2k-2})+\frac{1}{n+2k-1}\\
&>0.
\end{split}
\]

当 $p=2k$ 时,
\[
\begin{split}
(-1)^n(S_{n+p}-S_n)&=\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\frac{1}{n+4}+\cdots-\frac{1}{n+2k}\\
&=\frac{1}{n+1}-(\frac{1}{n+2}-\frac{1}{n+3})-\cdots-(\frac{1}{n+2k-2}-\frac{1}{n+2k-1})-\frac{1}{n+2k}\\
&\leqslant\frac{1}{n+1},
\end{split}
\]

\[
\begin{split}
(-1)^n(S_{n+p}-S_n)&=(\frac{1}{n+1}-\frac{1}{n+2})+(\frac{1}{n+3}-\frac{1}{n+4})+\cdots+(\frac{1}{n+2k-1}-\frac{1}{n+2k})\\
&>0.
\end{split}
\]

这说明
\[
|S_{n+p}-S_n|\leqslant\frac{1}{n+1}\rightarrow 0\quad(n\rightarrow\infty).\tag{*}
\]

因此原级数收敛.

在 $(*)$ 中令 $p\rightarrow\infty$, 则得 $|S-S_n|\leqslant\frac{1}{n+1}$, 其中 $S=\sum\limits_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}$ 为级数的和, 这是此交错级数的误差估计.