$s>0,\sum_{n=1}^\infty \frac{1}{n^{s+1}} < \frac{s+2}{s}$

hbghlyj

由于 $s+1>1$，伯努利不等式给出
\[
\Bigl(1 + \frac{1}{n}\Bigr)^{s+1} \ge 1 + \frac{s+1}{n}
\]
两边同除 $1+1/n$ 得
\[
\Bigl(1 + \frac{1}{n}\Bigr)^{-s} \le \frac{n + 1}{n + s + 1}
\]
于是
\[
1 - \Bigl(1 + \frac{1}{n}\Bigr)^{-s}
\ge 1 - \frac{n + 1}{n + s + 1}
= \frac{s}{n + s + 1}\ge\frac{s}{(s+2)n}
\]
并且
\[
\frac{1}{n^s} - \frac{1}{(n+1)^s}=\frac{1}{n^s} \Bigl(1 - \bigl(1 + \frac{1}{n}\bigr)^{-s}\Bigr)
\ge \frac{s}{(s+2)n^{s+1}}
\]
对 $n=1$ 到 $\infty$ 求和得
\[
\sum_{n=1}^\infty \frac{s}{(s+2)n^{s+1}} \le
1
\]
因此
\[
\sum_{n=1}^\infty \frac{1}{n^{s+1}} < \frac{s+2}{s}
\]

hbghlyj

对于 $0 < p < 1$，用伯努利不等式(截断二项式展开)
\[
(1-p) n^{-p} - \frac{p(1-p)}{2} n^{-p-1} < (n+1)^{1-p} - n^{1-p} < (1-p) n^{-p}.
\]
对 $n=1$ 到 $N$ 求和，利用$\sum_{n=1}^Nn^{-p-1}<\zeta(p+1)$得
\[
(1-p) \sum_{n=1}^N \frac{1}{n^p} - \frac{p(1-p)}{2}\zeta(p+1)<(N+1)^{1-p} - 1 < (1-p) \sum_{n=1}^N \frac{1}{n^p}
\]即
\[
\color{magenta}{\frac{(N+1)^{1-p} - 1}{1-p}}<\sum_{n=1}^N \frac{1}{n^p}<\color{magenta}{\frac{(N+1)^{1-p} - 1}{1-p}}+ \frac{p}{2} \zeta(p+1)
\]

hbghlyj

Cauchy condensation
For the upper bound, group the terms as follows (assuming $N \geq 1$ is an integer; the groups are from $n = N$ to $2N - 1$, $2N$ to $4N - 1$, $4N$ to $8N - 1$, and so on):
\[
T < \sum_{k=0}^\infty 2^k N \cdot \frac{1}{(2^k N)^p} = N^{1-p} \sum_{k=0}^\infty 2^{k(1-p)} = \frac{N^{1-p}}{1 - 2^{1-p}}.
\]
For the lower bound, consider groups from $n = N+1$ to $2N$, $2N+1$ to $4N$, and so on:
\[
T > \sum_{k=1}^\infty 2^{k-1} N \cdot \frac{1}{(2^k N)^p} = N^{1-p} \sum_{k=1}^\infty 2^{(k-1) - k p} = \frac{2^{-p} N^{1-p}}{1 - 2^{1-p}}.
\]