Prove-that-3-e-1-3-k-1-n-0-1-n-k2-k-

Question Number 161745 by HongKing last updated on 21/Dec/21

Prove that :

3 \sqrt{e} = \frac{1}{3} \underset{k = 1}{\sum^{\infty}} [\underset{n = 0}{\sum^{\infty}} \frac{1}{n!}] {k 2}^{- k}

Commented by mr W last updated on 22/Dec/21

q u e s t i o n s e e m s f a l s e .

Answered by mr W last updated on 22/Dec/21

\underset{k = 1}{\sum^{\infty}} x^{k} = \frac{x}{1 - x} f o r ∣ x ∣< 1

\underset{k = 1}{\sum^{\infty}} {k x}^{k - 1} = \frac{1}{1 - x} + \frac{x}{{(1 - x)}^{2}}

\underset{k = 1}{\sum^{\infty}} {k x}^{k} = \frac{x}{1 - x} + \frac{x^{2}}{{(1 - x)}^{2}}

w i t h x = \frac{1}{2}

\Rightarrow \underset{k = 1}{\sum^{\infty}} k 2^{- k} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} + \frac{{(\frac{1}{2})}^{2}}{{(1 - \frac{1}{2})}^{2}} = 2

e^{x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

w i t h x = 1

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} = e

\frac{1}{3} \underset{k = 1}{\sum^{\infty}} [\underset{n = 0}{\sum^{\infty}} \frac{1}{n!}] {k 2}^{- k}

= \frac{1}{3} \underset{k = 1}{\sum^{\infty}} (e) {k 2}^{- k}

= \frac{e}{3} \underset{k = 1}{\sum^{\infty}} {k 2}^{- k}

= \frac{e}{3} \times 2

= \frac{2 e}{3} \neq 3 \sqrt{e}

Commented by HongKing last updated on 22/Dec/21

SORRY my dear Sir, [\underset{n = 0}{\sum^{k}} \dots

no \infty, k