31 $sum_{kgeq 2}(zeta (k)-1)$
$=sum_{tgeq 2}sum_{kgeq 2}frac{1}{t^{k}}$
$=sum_{tgeq 2}frac{1}{(t-1)t}$
$=sum_{tgeq 2}(frac{1}{t-1}-frac{1}{t})=1$
$sum_{kgeq 1}(zeta (2k)-1)$
$=sum_{tgeq 2}sum_{kgeq 1}frac{1}{t^{2k}}$
$=sum_{tgeq 2}frac{1}{t^{2}-1}$
$=sum_{tgeq 2}frac{1}{2}(frac{1}{t-1}-frac{1}{t+1})$
$=frac{1}{2}(1+frac{1}{2})=frac{3}{4}$
32 分两种情况:
(1)$2nleq x < 2n+1$: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)$2n-1leq x < 2n$: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 _{kin K}a_{k}=infty $
$sum _{kin K}ca_{k}=csum _{kin K}a_{k}leftrightarrow Lambda _{kin K}(c+a_{k})=c+Lambda _{kin K}a_{k}$
$sum _{kin K}(a_{k}+b_{k})=sum _{kin K}a_{k}+sum _{kin K}b_{k}leftrightarrow Lambda _{kin K}min(a_{k},b_{k})=min(Lambda _{kin K}a_{k},Lambda _{kin K}b_{k})$
34 令$K^{+}=left { k|a_{k} geq 0 ight }$, $K^{-}=left { k|a_{k}<0 ight }$
对于任意奇数$n$，令$F_{n}=F_{n-1}igcup E_{n}$，其中$E_{n}subseteq K^{-}$且足够大，以致于$sum _{kin(F_{n-1}igcap K^{+)}}a_{k}+sum _{kin 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$1=sum _{kgeq 1}2^{-k}$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$frac{1}{2}=sum _{kgeq 1}3^{-k}$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$frac{1}{4}=sum _{kgeq 1}5^{-k}$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$g(g(g(n)))-g(g(g(n-1)))$
$=sum_{k=g(g(n-1))+1}^{g(g(n))}f(k)$
$=sum _{k}f(k)[g(g(n-1))<kleq g(g(n))]$
$=sum _{j,k}j[j=f(k)][g(g(n-1))<kleq g(g(n))]$
$=sum _{j,k}j[j=f(k)][g(n-1)<jleq g(n)]$
$=sum _{j,k}j(g(j)-g(j-1))[g(n-1)<jleq g(n)]$
$=sum _{j}jf(j)[g(n-1)<jleq g(n)]$
$=nsum _{j}j[g(n-1)<jleq g(n)]$
所以$g(g(g(n)))=1*1+(2+3)*2+(4+5)*3+(6+7+8)*4+(9+10+11)*5+...+(g(n)-g(n-1))n$
$=nsum_{k=1}^{g(n)}k-1*1-(1+2+3)*2-(1+2+3+4+5)*3-...-(1+2+..+g(n-1))*(n-1)$
$=frac{1}{2}ng(n)(g(n)+1)-frac{1}{2}sum_{k=1}^{n-1}g(k)(g(k)+1)$
37 应该能无限逼近。