Are you smarter than me? 6

The given series can be written as :

$\displaystyle\sum_{n=0}^∞ \frac{2^{n}}{n!} - 1$

The series of $e^{x}$ is :

$\displaystyle\sum_{n=0}^∞ \frac{x^{n}}{n!} = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +...$

Hence, the series so obtained is the one of $e^{x}$ with x=2 .

Therefore, our series becomes :

$\displaystyle\sum_{n=1}^∞ \frac{2^{n}}{n!} = e^{2} - 1 \approx \boxed{6.38}$

And hence the answer!

∑(1 to ∞) 2^n/n!= ∑(0 to ∞) (2^n/n!)-1 so the series of e^x is ∑(0 to ∞) x^n/n!=x^0/0! + x^1/1! +x^2/2! +x^3/3! + x^4/4! + x^5/5! +............... As the series obtained is the one of e^x with x=2 . So, our series becomes ... ∑(1 to ∞) 2^n/n! = e^2-1=7.38906-1=6.38906

∑(1 to ∞) 2^n/n!= ∑(0 to ∞) (2^n/n!)-1 so the series of e^x is ∑(0 to ∞) x^n/n!=x^0/0! + x^1/1! +x^2/2! +x^3/3! + x^4/4! + x^5/5! +............... As the series obtained is the one of e^x with x=2 . So, our series becomes ... ∑(1 to ∞) 2^n/n! = e^2-1=7.38906-1=6.38906