裂项放缩证明

$\displaystyle \sum^n_{k=1}\dfrac{2}{2k+1}\ln(n+1)<\ln(n+1)<\sum^n_{k=1}\dfrac{1}{\sqrt{k^2+k}}$
$\displaystyle \sum^n_{k=1}\dfrac{1}{\sqrt{2^k(2^k+1)}}>\ln \dfrac{2^{n+1}}{2^n+1}$
$\displaystyle \sum^n_{k=1}\dfrac{1}{k}>\ln(n+1)+\dfrac{1}{2}\dfrac{n}{n+1}$
$\displaystyle \sum^n_{k=1}\dfrac{2}{2k-1}-\ln(2n-1)\leq 2$
$\displaystyle \sum^n_{k=1}\dfrac{\ln k^2}{k^2}<\dfrac{(n-1)(2n+1)}{2(n+1)}$ , $n\geq 2$
$\displaystyle \sum^n_{k=2}\ln\dfrac{k-1}{k+1}>\dfrac{2-n-n^2}{\sqrt{2n(n+1)}}$
$\displaystyle \prod^n_{k=2}\dfrac{\ln k}{k+1}<\dfrac{1}{n}$
$(1+\dfrac{1}{n-1})^n>(1+\dfrac{1}{n})^{n+1}$ , $n\geq 2$
$\displaystyle \dfrac{n^{t+1}}{t+1}<\sum^n_{k=1} k^t<\dfrac{(n+1)^{t+1}}{t+1}$ ,$ t>0$

本作品采用知识共享署名-非商业性使用-禁止演绎 4.0 国际许可协议进行许可