陈洪葛的博客

For all postive real numbers $a_{1},a_{2},\cdots,a_{n}$,the following inequality holds
\[ \sum_{k=1}^{n}{a^{\frac{k}{k+1}}_{k}}\leq \sum_{k=1}^{n}{a_{k}}+\sqrt{\frac{2(\pi^2-3)}{9}\sum_{k=1}^{n}{a_{k}}} \]
\emph{Proof}:We begin with a preliminary result.\\
\textbf{Lemma}.If
\[ m_{k}=\left\{
     \begin{array}{ll}
       1, & \hbox{if \ $k=1$;} \\
       2\sqrt{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}, & \hbox{if \ $k>1$ .}
     \end{array}
   \right.
\]
then
\[ a_{k}^{\frac{k}{k+1}}\leq a_{k}+m_{k}\sqrt{a_{k}} \]
Proof,If $k=1$,this is trivially true.Otherwise,if $k>1$,the inequality can be written as
\[ a_{k}^{\frac{k}{k+1}}\left(1-a^{\frac{1}{k+1}}_{k} \right)\leq m_{k}\sqrt{a_{k}} \]
For the nontrivial case $0<a_{k}<1$,this inequality is equivalent to
\[ a^{\frac{k-1}{k+1}}_{k}\left(1-a^{\frac{1}{k+1}}_{k} \right)^{2}\leq m^{2}_{k} \]
To prove it,we use the AM-GM Inequality,as follows\\
\begin{align*}
a^{\frac{k-1}{k+1}}_{k}\left(1-a^{\frac{1}{k+1}}_{k}\right)^{2}
&=4(k-1)^{k-1}\left(\frac{a^{\frac{1}{k+1}}_{k}}{k-1}\right)^{k-1}\left(\frac{1-a^{\frac{1}{k+1}}_{k}}{2}\right)^{2}\\
&\leq4(k-1)^{k-1}\left[\frac{(k-1)\left(\frac{a^{\frac{1}{k+1}}_{k}}{k-1}\right)+2\left(\frac{1-a^{\frac{1}{k+1}}_{k}}{2}\right)}{k+1}\right]^{k+1}\\
&=\frac{4(k-1)^{k-1}}{(k+1)^{k+1}}\\
&=m^{2}_{k}
\end{align*}
and so the lemma is proved.\\
Returning to our problem,we see that it suffices to show that
\[ \sum_{k=1}^{n}{m_{k}\sqrt{a_{k}}}\leq \sqrt{\frac{2(\pi^2-3)}{9}\sum_{k=1}^{n}{a_{k}}}\]
But the \emph{Cauchy-Schwarz} Inequality gives us that
\[ \sum_{k=1}^{n}{m_{k}\sqrt{a_{k}}}\leq \sqrt{\left(\sum_{k=1}^{n}{m^{2}_{k}}\right)\left(\sum_{k=1}^{n}a_{k}\right)}\]
So,it remains to prove that
\[ \sum_{k=1}^{n}{m^{2}_{k}}\leq \frac{2(\pi^2-3)}{9}\]
or
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{\pi^2}{18}-\frac{1}{6}\]
Now,because
\[ \frac{(k+1)^{k+1}}{(k-1)^{k-1}}=(k+1)^{2}\left(\frac{k+1}{k-1}\right)^{k-1}=(k+1)^2\left(1+\frac{2}{k-1}\right)^{k-1}\]
\emph{Bernoulli's} Inequality yields
\[ \frac{(k+1)^{k+1}}{(k-1)^{k-1}}\geq 3(k+1)^{2} \]
and thus
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{1}{4}+\frac{1}{3}\sum_{k=2}^{n}{\frac{1}{(k+1)^{2}}}=\frac{1}{3}\sum_{k=1}^{n+1}{\frac{1}{k^2}}-\frac{1}{6}\]
On the other hand
\[ \lim_{n\rightarrow \infty}{\left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)}=\zeta{(2)}=\frac{\pi^2}{6}\]
and therefore it follows that
\[ \frac{1}{4}+\sum_{k=2}^{n}{\frac{(k-1)^{k-1}}{(k+1)^{k+1}}}\leq \frac{\pi^2}{18}-\frac{1}{6}\]
which finishes our proof.Notice also that equality does not occur.$\square$

设$f(x)\in C$是实值函数，且满足
\[ \int_{0}^{1}{f(x)dx}=\int_{0}^{1}{xf(x)dx}=\cdots=\int_{0}^{1}{x^{n-1}f(x)dx}=1 \]

(tian_275461)
证明:
\[ \int_{0}^{1}{(f(x))^{2}dx}\geq n^2 \]
证明:
首先，我们考虑多项式$P(x)$
\[ P(x)=a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}\]
若多项式$P(x)$也满足上面的条件，那么
\[ \int_{0}^{1}{(P(x))^{2}dx}=a_{0}+a_{1}+\cdots+a_{n-1} \]
为了求出系数$a_{i}$我们再次利用条件.
\[ \int_{0}^{1}{x^{k}P(x)dx}=1 \qquad k=0,1,\ldots n-1 \]
\[ \Rightarrow \frac{a_{0}}{k+1}+\frac{a_{1}}{k+2}+\cdots+\frac{a_{n-1}}{k+n}=1 \qquad  k=0,1,\ldots n-1 \]
设
\[ H(x)=\frac{a_{0}}{x+1}+\frac{a_{1}}{x+2}+\cdots+\frac{a_{n-1}}{x+n}-1 \]
则显然有
\[ H(0)=H(1)=\cdots=H(n-1)=0 \]
$H(x)$应该有
\[ H(x)=\frac{A\cdot x\cdot(x-1)\cdot(x-2)\cdots(x-n+1)}{(x+1)(x+2)\cdots(x+n)} \]
通过对比系数得$A=-1$,及
\[ a_{k}=(-1)^{n-k-1}\frac{(n+k)!}{(k!)^{2}\cdot(n-k-1)!}  \qquad  k=0,1,\ldots n-1 \]
用数学归纳法不难证明
\[ \sum_{k=0}^{n-1}{a_{k}}=n^{2} \]
所以，若多项式$P(x)$满足上面的性质，则
\[ \int_{0}^{1}{(P(x))^{2}dx}=a_{0}+a_{1}+\cdots+a_{n-1}=n^2 \]
取满足以上条件的多项式$P(x)$
应用Cauchy-Schwarz不等式
\[ \int_{0}^{1}{(P(x))^2dx}\int_{0}^{1}{(f(x))^2dx}\geq \left(\int_{0}^{1}{P(x)f(x)dx} \right)^{2}=n^4 \]
\[ \Rightarrow \int_{0}^{1}{(f(x))^{2}dx}\geq n^2 \]
Done!

求
\[ \lim_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}} \]
Solution:
\[  \sum_{k=1}^{n}\left(\frac{k}{n}\right)^{n}=\sum_{k=0}^{n-1}{\left(\frac{n-k}{n}\right)^{n}}=\sum_{k=0}^{n-1}{\left(1-\frac{k}{n}\right)^{n}} \]
\[ \left(1-\frac{k}{n}\right)^{n}=e^{n\ln(1-\frac{k}{n})}\leq e^{-k} \]
Thus.
\[ \varlimsup_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}}\leq\sum_{k=0}^{\infty}{e^{-k}}=\frac{e}{e-1} \]

for any $N\leq n-1$

\[  \sum_{k=0}^{N}{\left(\frac{n-k}{n}\right)^{n}}\leq \sum_{k=0}^{n-1}{\left(\frac{n-k}{n}\right)^{n}}\]

\[ \sum_{k=0}^{N}{\left(\frac{n-k}{n}\right)^{n}}\sim \sum_{k=0}^{N}e^{-k}=\frac{e^{-1}(1-e^{N+1})}{1-e^{-1}}\]

hence

\[ \varliminf_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}}\geq \frac{e^{-1}(1-e^{N+1})}{1-e^{-1}}\]

hold for any $N$,let $N\to \infty$,we have

\[ \lim_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}} =\frac{e}{e-1} \]
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事实上，这个问题可以加边成

\[ \lim_{n\to\infty}n\cdot\left(\frac{e}{e-1}-\sum_{i=1}^{n}\left(\frac{i}{n}\right)^{n}\right)=\frac{e(e+1)}{2(e-1)^3}\]
证明留给读者。

计算：

\[ \lim_{n\rightarrow\infty}{\left(1-\frac{1}{1\cdot 2}\right)\left(1-\frac{1}{2\cdot 3}\right)\cdots\left(1-\frac{1}{n(n+1)}\right)} \]

Solution：

We write

$ P=\prod_{n=1}^\infty \left(1-\frac{1}{n(n+1)}\right)=\prod_{n=1}^\infty \left(\frac{n^2+n-1}{n^2+n}\right)=\prod_{n=1}^\infty\frac{(n-a_1)(n-a_2)}{(n-b_1)(n-b_2)}  $,

where  $a_1=\frac{-1+\sqrt{5}}{2},\;a_2=\frac{-1-\sqrt{5}}{2},\;b_1=0$, and $b_2=-1.$

Using the [i]Weierstrass product[/i] for the gamma function, $\frac{1}{\Gamma(x)}=xe^\gamma x\prod_{n=1}^\infty \left(1+\frac{x}{n} \right)e^{-x/n}$, where $\gamma$ denotes Euler's constant, and the fundamental relation $\Gamma(x+1)=x\Gamma(x),$ we can deduce that

 $P= \prod_{j=1}^2 \frac{\Gamma(1-b_j)}{\Gamma(1-a_j)}=\frac{\Gamma(1)\Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right)\Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right)\Gamma\left(\frac{3+\sqrt{5}}{2}\right)}\approx 0.2966751356.$

[This standard method of determining certain infinite products is explained, for example, in Section 12.13 of [i]A Course of Modern Analysis [/i]([i]Fourth Edition[/i]) by E. T. Whittaker and G. N. Watson.]

Using the relation $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x},$ for $x$ not an integer, it should be possible to deduce a closed form for the infinite product given by the computer algebra system [i]Maple[/i]:  $\; - \frac{\sin \pi\left(\frac{1+\sqrt{5}}{2}\right)}{\pi}.$

\[ \int_{0}^{1}{\frac{\arctan{\sqrt{x^{2}+2}}}{(x^{2}+1)\sqrt{x^{2}+2}}dx} \]

(Ahmed's integral)

Solution:

\[ \frac{\pi^{2}}{16}=\int_{0}^{1}{\int_{0}^{1}{\frac{dxdy}{(1+x^2)(1+y^2)}}} \]
\[=\int_{0}^{1}{\int_{0}^{1}{\frac{1}{(1+x^{2})(2+x^2+y^2)}+\frac{1}{(1+y^{2})(2+x^2+y^2)}dxdy}}\]
\[=2\int_{0}^{1}{\int_{0}^{1}{\frac{1}{(1+x^2)(2+x^2+y^2)}dydx}} \]
\[=2\int_{0}^{1}{\frac{1}{(1+x^2)\sqrt{2+x^2}}\arctan{\frac{1}{\sqrt{2+x^2}}}dx} \]
\[ =2\int_{0}^{1}{\left(\frac{\pi}{2(1+x^2)\sqrt{2+x^2}}-\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}} \right)dx} \]
\[ =\frac{\pi^{2}}{6}-2\int_{0}^{1}{\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}}dx} \]
\[ \Rightarrow \int_{0}^{1}{\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}}dx}=\frac{5}{96}\pi^{2}\]