31 $\sum_{k\geq 2}(\zeta (k)-1)$
$=\sum_{t\geq 2}\sum_{k\geq 2}\frac{1}{t^{k}}$
$=\sum_{t\geq 2}\frac{1}{(t-1)t}$
$=\sum_{t\geq 2}(\frac{1}{t-1}-\frac{1}{t})=1$
$\sum_{k\geq 1}(\zeta (2k)-1)$
$=\sum_{t\geq 2}\sum_{k\geq 1}\frac{1}{t^{2k}}$
$=\sum_{t\geq 2}\frac{1}{t^{2}-1}$
$=\sum_{t\geq 2}\frac{1}{2}(\frac{1}{t-1}-\frac{1}{t+1})$
$=\frac{1}{2}(1+\frac{1}{2})=\frac{3}{4}$
32 分两种情况:
(1)$2n\leq x < 2n+1$:左边=$\sum_{k=0}^{n}k+\sum_{k=n+1}^{2n}(x-k)=\sum_{k=1}^{n}k+\sum_{k=1}^{n}(x-(k+n))=\sum_{k=1}^{n}(x-n)=n(x-n)$,右边等于$\sum_{k=0}^{n-1}(x-(2k+1))=nx-\sum_{k=0}^{n-1}(2k+1)=nx-n^{2}=n(x-n)$
(2)$2n-1\leq x < 2n$:左边=$\sum_{k=0}^{n-1}k+\sum_{k=n}^{2n-1}(x-k)=\sum_{k=0}^{n-1}k+\sum_{k=0}^{n-1}(x-(k+n))=\sum_{k=0}^{n-1}(x-n)=n(x-n)$,右边=$\sum_{k=0}^{n-1}(x-(2k+1))=n(x-n)$
33 首先假设如果$K$是空集，那么$\Lambda _{k\in K}a_{k}=\infty $
$\sum _{k\in K}ca_{k}=c\sum _{k\in K}a_{k}\leftrightarrow \Lambda _{k\in K}(c+a_{k})=c+\Lambda _{k\in K}a_{k}$
$\sum _{k\in K}(a_{k}+b_{k})=\sum _{k\in K}a_{k}+\sum _{k\in K}b_{k}\leftrightarrow \Lambda _{k\in K}min(a_{k},b_{k})=min(\Lambda _{k\in K}a_{k},\Lambda _{k\in K}b_{k})$
34 令$K^{+}=\left \{ k|a_{k} \geq 0 \right \}$, $K^{-}=\left \{ k|a_{k}<0 \right \}$
对于任意奇数$n$，令$F_{n}=F_{n-1}\bigcup E_{n}$，其中$E_{n}\subseteq K^{-}$且足够大，以致于$\sum _{k\in(F_{n-1}\bigcap K^{+)}}a_{k}+\sum _{k\in E_{n}}<A^{-} $
35 这里有一个巧妙的证明：
令$x=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+...$
（1）两边同时减去$1=\sum _{k\geq 1}2^{-k}$得到$x-1=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{9}+\frac{1}{10}+...$
（2）两边同时减去$\frac{1}{2}=\sum _{k\geq 1}3^{-k}$得到$x-1-\frac{1}{2}=1+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...$
（3）两边同时减去$\frac{1}{4}=\sum _{k\geq 1}5^{-k}$得到$x-1-\frac{1}{2}-\frac{1}{4}=1+\frac{1}{6}+\frac{1}{7}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...$
一直减下去得到$x-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{9}-...=1=\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{15}+...$
36 (1)根据定义有$g(n)-g(n-1)=f(n),g(0)=0$，所以$g(n)=\sum_{k=1}^{n}f(n)$
(2)由于$g(g(n))-g(g(n-1))=\sum_{k=g(n-1)+1}^{g(n)}f(k)=\sum_{k=g(n-1)+1}^{g(n)}n=n(g(n)-g(n-1))=nf(n)$
而$g(g(0))=0$，所以$g(g(n))=\sum_{k=1}^{n}kf(k)$
(3)由于$g(g(g(n)))-g(g(g(n-1)))$
$=\sum_{k=g(g(n-1))+1}^{g(g(n))}f(k)$
$=\sum _{k}f(k)[g(g(n-1))<k\leq g(g(n))]$
$=\sum _{j,k}j[j=f(k)][g(g(n-1))<k\leq g(g(n))]$
$=\sum _{j,k}j[j=f(k)][g(n-1)<j\leq g(n)]$
$=\sum _{j,k}j(g(j)-g(j-1))[g(n-1)<j\leq g(n)]$
$=\sum _{j}jf(j)[g(n-1)<j\leq g(n)]$
$=n\sum _{j}j[g(n-1)<j\leq g(n)]$
所以$g(g(g(n)))=1*1+(2+3)*2+(4+5)*3+(6+7+8)*4+(9+10+11)*5+...+(g(n)-g(n-1))n$
$=n\sum_{k=1}^{g(n)}k-1*1-(1+2+3)*2-(1+2+3+4+5)*3-...-(1+2+..+g(n-1))*(n-1)$
$=\frac{1}{2}ng(n)(g(n)+1)-\frac{1}{2}\sum_{k=1}^{n-1}g(k)(g(k)+1)$
37 应该能无限逼近。