Question

$\sf \: lim \: n \to \infty \: \: {n}^{ {n}^{ - 2} } {\bigg((n + 1)(n + \frac{1}{2} )(n + \frac{1}{ {2}^{2}} ) - - (n + \frac{1}{ {2}^{n - 1} } )}^{n}$
Evaluate the above limit.

Spammers please far away.

Quality answer needed.

Urgently required.

Please help.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm\displaystyle\lim_{n \to \infty } \: \: {n}^{ {n}^{ - 2} } {\bigg((n + 1)(n + \frac{1}{2} )(n + \frac{1}{ {2}^{2}} ) - - (n + \frac{1}{ {2}^{n - 1} } )\bigg)}^{n}$

can be rewritten as

$= \rm\displaystyle\lim_{n \to \infty } \frac{ {\bigg((n + 1)(n + \frac{1}{2} )(n + \frac{1}{ {2}^{2}} ) - - (n + \frac{1}{ {2}^{n - 1} } )\bigg)}^{n}}{ {( {n}^{n} )}^{2} }$

$\rm=\rm\displaystyle\lim_{n \to \infty } {\bigg[\dfrac{(n + 1)\bigg(n + \dfrac{1}{2}\bigg)\bigg(n + \dfrac{1}{ {2}^{2}}\bigg) - - - \bigg(n + \dfrac{1}{ {2}^{n - 1} } \bigg)}{ {n}^{n} } \bigg]}^{n}$

$\rm=\rm\displaystyle\lim_{n \to \infty } {\bigg[\dfrac{(n + 1)\bigg(n + \dfrac{1}{2}\bigg)\bigg(n + \dfrac{1}{ {2}^{2}}\bigg) - - - \bigg(n + \dfrac{1}{ {2}^{n - 1} } \bigg)}{ n \times n \times n \times - - - \times n } \bigg]}^{n}$

$\rm = \rm\displaystyle\lim_{n \to \infty } {\bigg[\bigg(1 + \dfrac{1}{n}\bigg)\bigg(1 + \dfrac{1}{2n}\bigg) - - - - \bigg(1 + \dfrac{1}{n {2}^{n - 1} }\bigg) \bigg]}^{n}$

$\rm = \rm\displaystyle\lim_{n \to \infty } {\bigg(1 + \dfrac{1}{n}\bigg)}^{n} {\bigg(1 + \dfrac{1}{2n}\bigg)}^{n} - - - - {\bigg(1 + \dfrac{1}{n {2}^{n - 1} }\bigg)}^{n}$

We know,

$\boxed{ \bf{ \: \rm\displaystyle\lim_{x \to \infty } {\bigg(1 + \dfrac{k}{x}\bigg)}^{x} = {e}^{k} }}$

So, using this identity, we get

$\rm \:  =  \:e \times {\bigg(e\bigg) }^{\dfrac{1}{2} } \times {\bigg(e\bigg) }^{\dfrac{1}{4} } \times - - - \times {\bigg(e\bigg) }^{\dfrac{1}{ {2}^{n - 1} } }$

$\rm \:  =  \: {\bigg(e\bigg) }^{1 + \dfrac{1}{2} + \dfrac{1}{4} + - - - + \dfrac{1}{ {2}^{n - 1} } }$

$\rm \:  =  \: {\bigg(e\bigg) }^{\dfrac{1\bigg[1 - \dfrac{1}{ {2}^{n} } \bigg]}{1 - \dfrac{1}{2} } }$

$\rm \:  =  \: {\bigg(e\bigg) }^{\dfrac{\bigg[1 - \dfrac{1}{ {2}^{n} } \bigg]}{ \dfrac{1}{2} } }$

$\rm \:  =  \: {\bigg(e\bigg) }^{2\bigg[1 - \dfrac{1}{ {2}^{n} } \bigg]}$