Question

$\red{ \left( \rm \displaystyle \rm \prod_{k = 0}^{ \sqrt{\pi} } \frac{1}{2} \left( \lim_{n\to\infty} e^{-n} \left( \sum_{k=0}^{n} \frac{n^k}{k!} \right) \right) \right)^{2} }$​

Answer 1

The expression can be simplified as follows:

$\displaystyle \rm \prod_{k = 0}^{ \sqrt{\pi} } \frac{1}{2} \left( \lim_{n\to\infty} e^{-n} \left( \sum_{k=0}^{n} \frac{n^k}{k!} \right) \right) = \left( \frac{1}{2} \lim_{n\to\infty} e^{-n} \left( \sum_{k=0}^{n} \frac{n^k}{k!} \right) \right)^{\sqrt{\pi}+1}$

Using the Taylor series expansion of the exponential function, we can write:

$\displaystyle \rm e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$

Substituting x = n, we get:

$\displaystyle \rm e^{n} = \sum_{k=0}^{\infty} \frac{n^k}{k!}$

Dividing both sides by eⁿ, we get:

$\displaystyle \rm \frac{1}{e^{n}} = \sum_{k=0}^{\infty} \frac{(-1)^k n^k}{k!}$

Substituting this expression in the original equation, we get:

$\displaystyle \rm \left( \frac{1}{2} \lim_{n\to\infty} \left( \sum_{k=0}^{n} \frac{(-1)^k n^k}{k!} \right) \right)^{\sqrt{\pi}+1}$

Using the Binomial Theorem, we can write:

$\displaystyle \rm \left( \frac{1}{2} \lim_{n\to\infty} \left( \sum_{k=0}^{n} \frac{(-1)^k n^k}{k!} \right) \right)^{\sqrt{\pi}+1} = \left( \frac{1}{2} \lim_{n\to\infty} \left( 1 - \frac{1}{n} \right)^n \right)^{\sqrt{\pi}+1}$

Taking the limit as n approaches infinity, we get:

$\displaystyle \rm \left( \frac{1}{2e} \right)^{\sqrt{\pi}+1}$

Raising this expression to the power of 2, we get:

$\displaystyle \rm \left( \frac{1}{2e} \right)^{2\sqrt{\pi}+2}$

Therefore, the value of the given expression is $\displaystyle \rm \left( \frac{1}{2e} \right)^{2\sqrt{\pi}+2}$.