Question-204991

Question Number 204991 by tigrecomplexe last updated on 04/Mar/24

Commented by tigrecomplexe last updated on 05/Mar/24

s o m m e o n e c a n p u t h a n d h e r e p l e a s e ?

Answered by witcher3 last updated on 05/Mar/24

u_{n} = {\underset{n \to \infty}{\lim e}}^{\sin (2 π \sqrt{1 + n^{2}}) \underset{k = 1}{\sum^{n}} \frac{\ln (1 + \frac{k}{n})}{2}}

\sin (2 π \sqrt{1 + n^{2}}) = \sin (2 π n {(1 + \frac{1}{n^{2}})}^{\frac{1}{2}})

{(1 + \frac{1}{n^{2}})}^{\frac{1}{2}} = 1 + \frac{1}{{2 n}^{2}} + o (\frac{1}{n^{2}})

\sin (2 π \sqrt{1 + n^{2}}) \overset{n \to \infty}{=} \sin (2 π n + \frac{π}{n} + o (\frac{1}{n})) = \sin (\frac{π}{n} + o (\frac{1}{n}))

\sin (2 π \sqrt{1 + n^{2}}) \overset{\infty}{=} \frac{π}{n} + o (\frac{1}{n})

u_{n} = e^{nsin (2 π \sqrt{1 + n^{2}}) . \frac{1}{2 n} \underset{k = 1}{\sum^{n}} \ln (1 + \frac{k}{n})}

\lim_{n \to \infty} . \frac{1}{2} . \frac{1}{n} \underset{k = 1}{\sum^{n}} \ln (1 + \frac{k}{n}) = \int_{0}^{1} \frac{\ln (1 + x)}{2} = {[\frac{(1 + x)}{2} \ln (1 + x) - \frac{x}{2}]}_{0}^{1} = \ln (2) - \frac{1}{2}

\lim_{n \to \infty} . nsin (2 π \sqrt{1 + n^{2}}) = \lim_{n \to \infty} n . (\frac{π}{n}) = π

U_{n} \to e^{π (\ln (2) - \frac{1}{2})} = \frac{2 π}{e^{\frac{π}{2}}}

Answered by Mathspace last updated on 05/Mar/24

A_{n} = {(\prod_{k = 1}^{n} \sqrt{1 + \frac{k}{n}})}^{s i n (2 π \sqrt{1 + n^{2}})}

\Rightarrow l o g (A_{n}) = s i n (2 π \sqrt{1 + n^{2})} \sum_{k = 1}^{n} l o g (\sqrt{1 + \frac{k}{n}})

\sqrt{1 + n^{2}} = n \sqrt{1 + \frac{1}{n^{2}}} = n {(1 + \frac{1}{n^{2}})}^{\frac{1}{2}}

\sim n (1 + \frac{1}{2 n^{2}}) = n + \frac{1}{2 n} \Rightarrow

s i n (2 π \sqrt{1 + n^{2}}) \sim s i n (2 π n + \frac{π}{n})

= s i n (\frac{π}{n}) \sim \frac{π}{n} \Rightarrow

l o g (A_{n}) \sim \frac{π}{n} \sum_{k = 1}^{n} l o g \sqrt{1 + \frac{k}{n}}

R e i m a n s u m g i v e

{l i m}_{n \to + \infty} l o g (A_{n})

= π \int_{0}^{1} l o g \sqrt{1 + x} d x

= \frac{π}{2} \int_{0}^{1} l o g (1 + x) d x b u t

\int_{0}^{1} l o g (1 + x) d x =

{[x l o g (1 + x)]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x} d x

= l o g 2 - \int_{0}^{1} \frac{1 + x - 1}{1 + x} d x

= l o g 2 - 1 + \int_{0}^{1} \frac{d x}{1 + x}

= l o g 2 - 1 + {[l o g (1 + x)]}_{0}^{1}

= l o g 2 - 1 + l o g 2 = 2 l o g 2 - 1

l i m l o g (A_{n}) = \frac{π}{2} (2 l o g 2 - 1)

= π l o g 2 - \frac{π}{2} \Rightarrow

{l i m}_{n \to + \infty} A_{n} = e^{π l o g 2 - \frac{π}{2}}

= 2^{π} \times e^{- \frac{π}{2}}