อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Mastermander
140.
$$0<\lim_{n\to\infty} \dfrac{\sqrt[n]{n!}}{n}<1$$
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True...
Let $\displaystyle{\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}=x}$
$\displaystyle{\ln x=\lim_{n \rightarrow \infty} \frac{1}{n} \ln \left ( \frac{n!}{n^n} \right )=\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln \left ( \frac{k}{n} \right )= \int_0^1 \ln x dx=-1 \rightarrow x= \frac{1}{e} }$
141.$\displaystyle{\int_0^\infty\ln\left(x+\frac{1}{x}\right)\frac{dx}{1+x^2}=\int_0^\frac{\pi}{2}\left(\frac{\theta}{\sin\theta }\right)^2d\theta}$
142.$\displaystyle{\sum_{n=1}^\infty\frac{1}{n^2+\sqrt{n}}=\frac{\pi}{e}}$
143.$\displaystyle{\sum_{n=0}^\infty\frac{n^n}{(n+1)^{n+1}}}$...ลู่เข้า
__________________
$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$
BUT
$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$