Prove that t n = 1 + ln ⁡ ( 2 n ) 2 n + o ( 1 n ) as n tends to infinity.

Question

Prove that       t          n        =  1  +            ln      ⁡      (      2      n      )              2      n        +  o      (          1      n        )   as n tends to infinity.

tennispopj8 · Accepted Answer

For (iii) (which proves (ii) as well): First note that

t_{n}^{2 n} + 1 = 2 n t_{n} \geq 2 n,

i.e.,

t_{n} \geq \sqrt[2 n]{2 n - 1} .

But

\sqrt[2 n]{2 n - 1} = \exp (\frac{\log (2 n - 1)}{2 n}) = 1 + \frac{\log (2 n - 1)}{2 n} + o (\frac{1}{n}) = 1 + \frac{\log (2 n)}{2 n} + o (\frac{1}{n}) .

Thus,

\begin{matrix} (1) & t_{n} \geq 1 + \frac{\log (2 n)}{2 n} + o (\frac{1}{n}) . \end{matrix}

We also have

2 n t_{n} = t_{n}^{2 n} + 1 \geq t_{n}^{2 n},

i.e.,

\sqrt[2 n - 1]{2 n} \geq t_{n} .

But

\sqrt[2 n - 1]{2 n} = \exp (\frac{\log (2 n)}{2 n - 1}) = 1 + \frac{\log (2 n)}{2 n - 1} + o (\frac{1}{n}) = 1 + \frac{\log (2 n)}{2 n} + o (\frac{1}{n}) .

\begin{matrix} (2) & t_{n} \leq 1 + \frac{\log (2 n)}{2 n} + o (\frac{1}{n}) . \end{matrix}

From (1) and (2),

t_{n} = 1 + \frac{\log (2 n)}{2 n} + o (\frac{1}{n}),

as desired.

Prove that t <mrow class="MJX-TeXAtom-ORD"> n </mrow> </msub> = 1 +

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