Question

Calculus:

(See photo for question)

I can't seem to get the answer because the factorials are throwing me off. My numerator should be good to go, however. It's really just that denominator that's messing-up my final answer.

Answer 1

\[S = \sum_{n=1}^\infty\dfrac{2^{n+2}}{3\cdot n!}\\~\\~\\3S = \sum_{n=1}^\infty\dfrac{2^{n+2}}{n!}\\~\\~\\\dfrac{3}{4}S = \sum_{n=1}^\infty\dfrac{2^n}{n!}\]Does that help you?

Answer 2

Not exactly, but I found out that the answer is \[\frac{ 4 }{ 3 }(e^2-1)\]

I'm still retracing the steps on how to get it.

Answer 3

Are you familiar with Taylor series?

Answer 4

Have you considered the fact that \(\displaystyle e^x = \sum_{n=0}^\infty\dfrac{x^n}{n!}\) ?

Answer 5

Yes I've done all that.

Answer 6

So you figured out?

Answer 7

Sort of... I asked another website for help and got the correct answer, but the interpretation is still percolating in my mind.