why $\sum_{n=2}^{+\infty}\frac{(2)^n}{n!} = e^{-2} - 3 $

Question

please how to show it ? I need to find the exact sum of :
$\sum_{n=2}^{+\infty}\frac{(2)^n}{n!} = e^{-2} - 3 $

Answer 1

Consider the Taylor series expansion $e^x=\sum_{i=0}^{\infty}\frac{x^n}{n!}.$

So at $x=2,\quad e^2=\sum_{n=0}^{\infty}\frac{2^n}{n!}=3+\sum_{n=2}^{\infty}\frac{2^n}{n!}\implies \sum_{n=2}^{\infty}\frac{2^n}{n!}=e^2-3.$

Answer 2

HINT

Try to use the Taylor series expansion of the exponential which is given by

$$\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\sum_{n=2}^{\infty}\frac{x^n}{n!}$$

Answer 3

Hint: $$ a+b = c+d $$ implies $$ b = c+d -a $$ Can you take it from here?

Answer 4

You wanted $$\sum_{n=2}^{+\infty}\frac{(2)^n}{n!} = e^{2} - 3$$

You need to know the Taylor series for $e^x$, namely

$$e^x = \sum_{n=0}^{+\infty}\frac{(x)^n}{n!}=1+x+\sum_{n=2}^{+\infty}\frac{(x)^n}{n!}$$

All you have to do from here is to let $x=2$ and subtract the first two terms.

Answer 5

Note that

$$\sum_{n=2}^{\infty}\frac{(2)^n}{n!} >\sum_{n=2}^{2}\frac{(2)^n}{n!} = 2> e^{-2} -3$$

therefore the given equality cannot be true.