Question

2/3!+4/5!+6/7!+....to infinity is equal to?

Answer 1

What you've got there is a basic series...\[\sum_{n=0}^{\infty}\frac{n+2}{n!}\]Maybe you should try using the ratio test. That's what I'd first use to see if it converges.

Answer 2

the series notation is wrong

Answer 3

i dont think so buddy

Answer 4

Sorry, the sum's off. :P Don't have time to work it out right now, but I'm interested to see if anyone's got it to a concise form

Answer 5

m with you..;-)

Answer 6

1/2 is the answer?

Answer 7

no its 1/e

Answer 8

which book r u using?

Answer 9

"rudiments of mathematics" its a book for wbjee(west bengal joint entnc xm)

Answer 10

Wow, I wasn't even thinking. Try this\[\sum_{n=1}^{\infty}{\frac{2n}{(2n+1)!}}\]

Answer 11

good!

Answer 12

(2n+2)/(2(n+3) *(2n+1)!/2n

Answer 13

(2n+2)/(2n+3)! *(2n+1)!/2n

Answer 14

wow so u got the expression right finally

Answer 15

\[2n+2 \over {(2n+2)(2n+3)(2n)}\]

Answer 16

\[\lim n \rightarrow \infty [1/2n(2n+3)\]

Answer 17

zero

Answer 18

This just tell us that it converge but not to what

Answer 19

yup,convergent

Answer 20

1/e= e^-1[1+1/2! +1/3!...]^-1
that can be expanded using binomial expension

Answer 21

\[[1+(1/2/!+1/3!...)]^{-1}=1- (1/2!+1/3!...)+(1/2!+ 1/3!...)^2....\]

Answer 22

how come the sum is 2/3!+4/5!+6/7!+....?

Answer 23

i dont kno. thats what the book says.. its wrong isn't it?

Answer 24

seems to be, not sure

Answer 25

wow..great thinkin there buddy.. thanx!