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Question Number 217088 by alcohol last updated on 28/Feb/25

s h o w t h a t \int_{n}^{n + 1} l n (t) d t ⩽ l n (n + \frac{1}{2})

G i v e n u_{n} = \frac{{(4 n)}^{n} n! e^{- n}}{(2 n)!}, \forall n ⩾ 1

p r o v e, u s i n g t h e p r e c e d i n g q u e s t i o n t h a t

u_{n} i s d e c r e a s i n g a n d c o n v e r g e n t

Answered by MrGaster last updated on 01/Mar/25

Prove : \int_{n}^{n + 1} \ln (t) d t \leq \ln (n + \frac{1}{2})

\int_{n}^{n + 1} \ln (t) d t = {[t \ln (t) - t]}_{n}^{n + 1}

= ((n + 1) + \ln (n + 1) - (n + 1) - (n \ln (n) - n)

= (n + 1) \ln (n + 1) - n \ln (n) - 1

\ln (n + 1) \leq \ln (n) + \frac{1}{n} ⊢ (n + 1) \ln (n + 1) \leq (n + 1) (\ln (n) + \frac{1}{n})

= (n + 1) \ln (n) + (n + 1) \frac{1}{n}

= (n + 1) \ln (n) + 1 \frac{1}{n}

∴ (n + 1) \ln (n + 1) - n \ln (n) - 1 \leq ((n + 1) \ln (n) + 1 + \frac{1}{n}) - n \ln (n) - 1

= \ln (n) + \frac{1}{n}

∵ \frac{1}{n} \leq \ln (1 + \frac{1}{2 n}) \Rightarrow \ln (n) + \frac{1}{n} \leq \ln (n) + \ln (1 + \frac{1}{2 n})

= \ln (n (1 + \frac{1}{2 n}))

= \ln (n + \frac{1}{2})

so, \begin{matrix} \int_{n}^{n + 1} \ln (t) d t \leq \ln (n + \frac{1}{2}) \end{matrix}

Give u_{n} = \frac{{(4 n)}^{n} n! e^{- n}}{(2 n)!}, \forall n \geq 1, Prove that u_{n} is decreasing and converhgent .

Part 1 : Prove u_{n} is decreasing .

Consider the ratio :

\frac{u_{n + 1}}{u_{n}} = \frac{\frac{(4 {(n + 1)}^{n + 1} (n + 1)! e^{- (n + 1)}}{(2 (n + 1)!}}{\frac{{(4 n)}^{n} e^{- n}}{(2 n)!}}

= \frac{{(4 (n + 1))}^{n + 1} (n + 1) e^{- 1}}{{(4 n)}^{n} (2 n + 2) (2 n + 1)}

= \frac{4^{n + 1} {(n + 1)}^{n + 1} (n + 1) e^{- 1}}{4^{n} n^{n} (2 n + 2) (2 n + 1)}

= \frac{4 {(n + 1)}^{n + 2} e^{- 1}}{n^{n} 2 (n + 1) (2 n + 1)}

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

∵ {(n + 1)}^{n + 1} > n^{n + 1} \Rightarrow \frac{{(n + 1)}^{n - 1}}{n^{n}} > 1

∴ \frac{u_{n + 1}}{u_{n}} < 1 for large n, implying u_{n} is decreasing .

Step 2 : Prove u_{n} is convergent . Using {stirling}^{'} s approximation n! \approx \sqrt{2 π n} {(\frac{n}{e})}^{n} \Rightarrow u_{n} = \frac{{(4 n)}^{n} n! e^{- n}}{(2 n)!} \approx \frac{{(4 n)}^{n} \sqrt{2 π n} {(\frac{n}{e})}^{n} e^{- n}}{\sqrt{4 π n} {(\frac{2 n}{e})}^{2 n}}

= \frac{{(4 n)}^{n} \sqrt{2 π n} n^{n} e^{- 2 n}}{\sqrt{4 π n} 2^{2 n} n^{2 n}}

= \frac{\sqrt{2 π n} 4^{n} n^{n} e^{- 2 n}}{\sqrt{4 π n} 2^{n} n^{2 n}}

= \frac{\sqrt{2 π} 4^{n} e^{- 2 n}}{\sqrt{4 π} 2^{2 n} n^{n}}

= \frac{\sqrt{2} 2^{2 n} e^{- 2 n}}{\sqrt{2} 2^{2 n} n^{n}}

= \frac{e^{2 n}}{n^{n}}

As n \to \infty, e^{- 2 n} \to 0 and n^{n} \to \infty, so u_{n} \to 0 . Thus, u_{n} is convergent .

\begin{matrix} u_{n} is decreasing and convergent \end{matrix}

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