Question

نشان دهید: $$ \lim_{n\to \infty } [tan( \gamma -H_n+ \frac{ \pi }{4} +log(n)]^{ \frac{1}{sin( \gamma -H_n+log(n))} }= e^{2} $$

Comments

$$f(n)= [tan ( \gamma - H_{n} + \frac{ \pi }{4} +log n)]^{ \frac{1}{sin( \gamma - H_{n} +logn)} } \Longrightarrow lnf(n)= \frac{ln[tan( \gamma -H_n+ \frac{ \pi }{4} +logn)]}{sin( \gamma - H_{n} +logn)}  \wedge  \ast  \frac{d}{dn} H_{n} = \frac{d}{dn} [ \psi (n+1)+ \gamma ]= \psi '(n+1) \wedge  \psi (n+1)= \frac{1}{n}  + \psi (n) \Longrightarrow  \psi '(n+1)= \psi '(n)- \frac{1}{ n^{2} }=- \frac{1}{ n^{2} }  - \frac{1}{ (n-1)^{2} } + \psi (n-1) = -\frac{1}{ n^{2} }- \frac{1}{ (n-1)^{2} } -...- \frac{1}{ 1^{2} } + \psi '(1)=- H''_{n}  + \psi '(1) \wedge  \ast  \psi '(1)=- \int _0^1 \frac{lnx}{1-x} dx  \Longrightarrow  \psi (n+1)=- H''_{n}- \int _0^1 \frac{lnx}{1-x} dx =- H''_{n} + \zeta (2) \Longrightarrow  \frac{a}{b} [ln(tan ( \gamma - H_{n}+ \frac{ \pi }{4}+logn))]= \frac{ sec^{2} ( \gamma - H_{n}+ \frac{ \pi }{4} +logn) }{tan( \gamma - H_{n} + \frac{ \pi }{4} +logn)} .( \frac{1}{n} + H''_{n} - \zeta (2))  \wedge  \ast  \frac{d}{dn}sin ( \gamma - H_{n}+logn)=cos( \gamma - H_{n}+logn)( \frac{a}{b} + H''_{n}- \zeta (2))  \Longrightarrow L'= \lim_{n\to  \infty }  \frac{ sec^{2}( \gamma - H_{n}+ \frac{ \pi }{4} +logn)  }{tan( \gamma - H_{n} + \frac{ \pi }{4}+logn) } . \frac{ \frac{1}{n}+ H''_{n}- \zeta (2)  }{cos( \gamma  -H_{n} +logn)( \frac{1}{n} + H''_{n}- \zeta (2)) } = \frac{ sec^{2}  \frac{ \pi }{4} }{tan \frac{ \pi }{4} }=2 \wedge  \ast  \lim_{n\to  \infty }(logn- H_{n} )=- \gamma  \Longrightarrow L= e^{2} $$

Answer 1

$tan(x+ \frac{ \pi }{4} )= \frac{tanx+tan(\frac{ \pi }{4})}{1-tanx.tan(\frac{ \pi }{4})}= \frac{1+tanx}{1-tanx} $

حالا قرار دهید:

$x_n:= \gamma -(H_n-log(n)) \Rightarrow \lim_{n\to \infty } x_n=0$(چرا؟)

$ \Rightarrow \lim_{n\to \infty } (1+tanx_n)^ \frac{1}{sinx_n}=\lim_{n\to \infty } [(1+tanx_n)^ \frac{cosx_n}{sinx_n}]^{ \frac{1}{cosx_n} }$

$=\lim_{n\to \infty } [(1+tanx_n)^ \frac{1}{tanx_n}]^{ \frac{1}{cosx_n} }= \lim_{n\to \infty } e^{\frac{1}{cosx_n}}=e^1=e$(چرا؟)

$, \lim_{n\to \infty } (1-tanx_n)^ \frac{1}{sinx_n}=\lim_{n\to \infty } [(1-tanx_n)^ \frac{-cosx_n}{sinx_n}]^{\frac{-1}{cosx_n} }$

$=\lim_{n\to \infty } [(1-tanx_n)^ \frac{-1}{tanx_n}]^{ \frac{-1}{cosx_n} }= \lim_{n\to \infty } e^{-\frac{1}{cosx_n}}=e^{-1} $(چرا؟)

$ \Rightarrow \lim_{n\to \infty }[tan( \gamma -H_n+ \frac{ \pi }{4} +log(n))]^ \frac{1}{sin( \gamma -H_n+log(n))}= \frac{\lim_{n\to \infty } (1+tanx_n)^ \frac{1}{sinx_n}}{\lim_{n\to \infty } (1-tanx_n)^ \frac{1}{sinx_n}}= \frac{e}{e^{-1}} =e^2$

$ \Box $