交错调和级数变式求问
原题为\(\displaystyle \int_0^1 \left[\frac{n}{x}\right]-n\left[\frac{1}{x}\right] \dif x=\dots= n\sum_{q=1}^{\infty} \frac{1}{1+nq}+\frac{1}{2+nq}+\dots+\frac{1}{n-1+nq}-\frac{n-1}{n+nq}=n\left(\ln n-\sum_{k=2}^n \frac{1}{k}\right) \)求证\(\displaystyle \sum_{q=1}^{\infty} \frac{1}{1+nq}+\frac{1}{2+nq}+\dots+\frac{1}{n-1+nq}-\frac{n-1}{n+nq}=\ln n-\sum_{k=2}^n \frac{1}{k} \) 反而被积分证了
\(\displaystyle \int_{\frac{1}{u}}^1 \left[\frac{n}{x}\right] \dif x=\int_{\frac{n}{n+1}}^{\frac{n}{n}} n \dif x+\int_{\frac{n}{n+2}}^{\frac{n}{n+1}} (n+1) \dif x +\dots+\int_{\frac{n}{nu}}^{\frac{n}{nu-1}} (nu-1) \dif x=\sum_{k=n}^{nu-1} k\left(\frac{n}{k}-\frac{n}{k+1}\right)=n\sum_{k=n}^{nu-1} \frac{1}{k+1}\)
\(\displaystyle \int_{\frac{1}{u}}^1 n\left[\frac{1}{x}\right] \dif x=n\int_{\frac{1}{2}}^{1} 1 \dif x+\int_{\frac{1}{3}}^{\frac{1}{2}} 2 \dif x +\dots+\int_{\frac{1}{u}}^{\frac{1}{u-1}} (u-1) \dif x=n\sum_{k=1}^{u-1} k\left(\frac{1}{k}-\frac{1}{k+1}\right)=n\sum_{k=1}^{u-1} \frac{1}{k+1}\)
\(\displaystyle \int_0^1 \left[\frac{n}{x}\right]-n\left[\frac{1}{x}\right] \dif x=n\sum_{k=n}^{nu-1} \frac{1}{k+1}-n\sum_{k=1}^{u-1} \frac{1}{k+1}=n\left(\sum_{k=u}^{nu-1} \frac{1}{k+1}-\sum_{k=1}^{n-1} \frac{1}{k+1}\right) \rightarrow n\left(\ln n-\sum_{k=2}^n \frac{1}{k}\right) \) 移过去
\(\displaystyle \sum_{q=0}^{\infty} \frac{1}{1+nq}+\frac{1}{2+nq}+\dots+\frac{1}{n-1+nq}-\frac{n-1}{n+nq}=\ln n \) 哦
\(\displaystyle \sum_{k=0}^u \frac{1}{1+nk}+\frac{1}{2+nk}+\dots+\frac{1}{n+nk}-\frac{1}{1+k}=\sum_{k=1}^{nu+1} \frac{1}{k}-\sum_{k=1}^{u+1} \frac{1}{k}=\sum_{k=n+2}^{nu+1} \frac{1}{k} \rightarrow \ln n\)
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