求和的一些估计式
计算$$\displaystyle\lim_{n\rightarrow\infty}\left({2\sqrt n}-\sum_{k=1}^n\frac{1}{\sqrt k}\right).$$
Use $\sqrt{n} = \sum_{k=1}^n \left( \sqrt{k} - \sqrt{k-1} \right)$, then
\begin{align*}&2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} \\=& \sum_{k=1}^n \left( 2 \sqrt{k} - 2 \sqrt{k-1} - \frac{1}{\sqrt{k}} \right) = \sum_{k=1}^n \frac{1}{\sqrt{k}} \left( \sqrt{k}-\sqrt{k-1} \right)^2\\=& \sum_{k=1}^n \frac{1}{\sqrt{k}} \left( \frac{(\sqrt{k}-\sqrt{k-1})(\sqrt{k}+\sqrt{k-1})}{(\sqrt{k}+\sqrt{k-1})} \right)^2 \\=& \sum_{k=1}^n \frac{1}{\sqrt{k} \left(\sqrt{k}+\sqrt{k-1}\right)^2}.\end{align*}
This shows the limit does exist and $\lim_{n \to \infty} \left( 2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} \right) = \sum_{k=1}^\infty \frac{1}{\sqrt{k} \left(\sqrt{k}+\sqrt{k-1}\right)^2}$.
The value of this sums equals $-\zeta\left(\frac{1}{2} \right) \approx 1.4603545$. This value is found by other means, though:
$$2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} = 2 \sqrt{n} - \left( \zeta\left(\frac{1}{2}\right) - \zeta\left(\frac{1}{2}, n+1\right)\right) \sim -\zeta\left(\frac{1}{2}\right) - \frac{1}{2\sqrt{n}} + o\left( \frac{1}{n} \right) .$$
取整函数的求和问题
证明\[\sum\limits_{k = 1}^{n - 1} {\sin \left( {\frac{{\left( {2\left\lfloor {\sqrt {kn} } \right\rfloor + 1} \right)\pi }}{{2n}}} \right)} = \cot \left( {\frac{\pi }{{2n}}} \right)\cos \left( {\frac{\pi }{{2n}}} \right),\]
where $\lfloor\cdot\rfloor$ is greatest integer function.
Lemma Summation by Pasts (1)http://en.wikipedia.org/wiki/Summation_by_parts
$$\sum_{k=a}^b f_k\Delta g_k=f_kg_k\Bigg|_{k=a}^{b+1}-\sum_{k=a}^b g_{k+1}\Delta f_k=f_{b+1}g_{b+1}-f_ag_a-\sum_{k=a}^b g_{k+1}\Delta f_k$$
$\displaystyle \text{with difference operator }\Delta :\qquad \Delta f_k=f_{k+1}-f_k$.
My solution
$\text{Get }j=\left\lfloor \sqrt{kn}\right\rfloor\Rightarrow j\le \sqrt{kn}<j+1\Rightarrow \dfrac{j^2}{n}\le k<\dfrac{(j+1)^2}{n}$
Therefore if $\left\lfloor\dfrac{j^2}{n}\right\rfloor+1\le k\le \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor$ then $\sqrt{kn}=j$
Note:
$j=0\Rightarrow \left\lfloor\dfrac{j^2}{n}\right\rfloor+1=1$
$j=n-2\Rightarrow \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor=n-2$
$j=n-1\Rightarrow \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor=n>n-1$
So that sum became
\begin{align*}S&=\sum_{k=1}^{n-1}\sin\left(\dfrac{\left(2\lfloor\sqrt{kn}\rfloor+1\right)\pi}{2n}\right)\\&=\sin\left(\dfrac{\left(2\lfloor\sqrt{(n-1)n}\rfloor+1\right)\pi}{2n}\right)+\sum_{k=1}^{n-2}\sin\left(\dfrac{\left(2\lfloor\sqrt{kn}\rfloor+1\right)\pi}{2n}\right)\\ &=\sin\left(\frac{(2n-1)\pi}{2n}\right)+\sum_{j=0}^{n-2}\left(\left\lfloor\frac{(j+1)^2}{n}\right\rfloor-\left\lfloor\frac{j^2}{n}\right\rfloor\right)\sin\left(\frac{(2j+1)\pi}{2n}\right)\\ &=\sin\left(\frac{\pi}{2n}\right)+\sum_{j=0}^{n-2}\Delta\left(\left\lfloor\frac{j^2}{n}\right\rfloor\right)\sin\left(\frac{(2j+1)\pi}{2n}\right)\end{align*}
Using the [1], we get
\begin{align*}S&=\sin\left(\frac{\pi}{2n}\right)+\left[\sin\left(\frac{(2j+1)\pi}{2n}\right)\left\lfloor\frac{j^2}{n}\right\rfloor\right]_{j=0}^{n-1}\\ &{}\quad -\sum_{j=0}^{n-2}\left\lfloor\frac{(j+1)^2}{n}\right\rfloor\left(\sin\left(\frac{(2j+3)\pi}{2n}\right)-\sin\left(\frac{(2j+1)\pi}{2n}\right)\right)\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)\sum_{j=0}^{n-2}\left\lfloor\frac{(j+1)^2}{n}\right\rfloor\cos\left(\frac{(j+1)\pi}{n}\right)\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)\underbrace{\sum_{j=1}^{n-1}\left\lfloor\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)}_{=A}\end{align*}
With sum A, using reverse summand property we get
\begin{align*}A&=\sum_{j=1}^{n-1}\left\lfloor\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)\\&=\sum_{j=1}^{n-1}\left\lfloor\frac{(n-j)^2}{n}\right\rfloor\cos\left(\frac{(n-j)\pi}{n}\right)\\ &=-\sum_{j=1}^{n-1}\left\lfloor n-2j+\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)\\ &=-A+\sum_{j=1}^{n-1}(2j-n)\cos\left(\frac{j\pi}{n}\right)\\ \Rightarrow A&=\frac{1}{2}\sum_{j=1}^{n-1}(2j-n)\cos\left(\frac{j\pi}{n}\right)\end{align*}
We get
$\displaystyle\cos\left(\frac{j\pi}{n}\right)=\dfrac{1}{2\sin\left(\frac{\pi}{2n}\right)}\left[\sin\left(\frac{(2j+1)\pi}{2n}\right)-\sin\left(\frac{(2j-1)\pi}{2n}\right)\right]=\dfrac{1}{2\sin\left(\frac{\pi}{2n}\right)}\Delta\left[\sin\left(\frac{(2j-1)\pi}{2n}\right)\right]$
continue using [1] :)
$\displaystyle A =\left.\dfrac{(2j-n)}{4\sin\left(\frac{\pi}{2n}\right)}\sin\left(\frac{(2j-1)\pi}{2n}\right)\right|_{j=1}^{n}-\dfrac{1}{4\sin\left(\frac{\pi}{2n}\right)}\sum_{j=1}^{n-1}2\sin\left(\frac{(2j+1)\pi}{2n}\right)$
\begin{align*}A&=\frac{n-1}{2}+\dfrac{1}{4\sin^2\left(\frac{\pi}{2n}\right)}\sum_{j=1}^{n-1}\Delta\left[\cos\left(\frac{j\pi}{n}\right)\right]\\ &=\frac{n-1}{2}+\dfrac{1}{4\sin^2\left(\frac{\pi}{2n}\right)}\cdot\left.\cos\left(\frac{j\pi}{n}\right)\right|_{j=1}^n\\ &=\frac{n-1}{2}-\dfrac{1+\cos\left(\frac{\pi}{n}\right)}{4\sin^2\left(\frac{\pi}{2n}\right)}\\&=\frac{n-1}{2}-\dfrac{2\cos^2\left(\frac{\pi}{2n}\right)}{4\sin^2\left(\frac{\pi}{2n}\right)}\end{align*}
Therefore:
\begin{align*}S&=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)A\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-(n-1)\sin\left(\frac{\pi}{2n}\right)+\dfrac{\cos^2\left(\frac{\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}\\ &=\boxed{\displaystyle\cot\left(\frac{\pi}{2n}\right)\cos\left(\frac{\pi}{2n}\right)}\end{align*}
