Question

then I did a couple of these questions

Answer 1

the first one:
\[\large \sum_{n=0}^{\infty}~(-1)^n\frac{2^n}{n!}\]re-writing:
\[\large \sum_{n=0}^{\infty}\frac{(-2)^n}{n!}\]then I can use the fact that:
\[\large e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\]
and thus\[\large \sum_{n=0}^{\infty}~(-1)^n\frac{2^n}{n!}=e^{-2}=1/e^2\]

Answer 2

excuse me ?

Answer 3

what was the question

Answer 4

sorry

Answer 5

what this series of (-1)^n2^n/n! converges to

Answer 6

checking, one moment

Answer 7

take your time

Answer 8

nice pic, btw

Answer 9

thanks ☺

Answer 10

you did it right, your logic is correct

Answer 11

tnx for verification!

Answer 12

$$\Large e^{-2}=\sum_{n=0}^{\infty}\frac{(-2)^n}{n!}


$$

Answer 13

yeah

Answer 14

and so would be for any C in the place of -2, (that it would be e^C

Answer 15

which is e^x

Answer 16

$$\Large{ e^{-2}
=\sum_{n=0}^{\infty}\frac{(-2)^n}{n!}

\\~\\=\sum_{n=0}^{\infty}\frac{(~(-1)\cdot 2)^n}{n!}
\\~\\=\sum_{n=0}^{\infty}\frac{(-1)^n(2)^n}{n!}
= \sum_{n=0}^{\infty}~(-1)^n\frac{2^n}{n!}
}
$$

Answer 17

correct ☺

Answer 18

yes.... got it

Answer 19

tnx again