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

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

Mathematics
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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.
the series notation is wrong
i dont think so buddy

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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
m with you..;-)
1/2 is the answer?
no its 1/e
which book r u using?
"rudiments of mathematics" its a book for wbjee(west bengal joint entnc xm)
Wow, I wasn't even thinking. Try this\[\sum_{n=1}^{\infty}{\frac{2n}{(2n+1)!}}\]
good!
(2n+2)/(2(n+3) *(2n+1)!/2n
(2n+2)/(2n+3)! *(2n+1)!/2n
wow so u got the expression right finally
\[2n+2 \over {(2n+2)(2n+3)(2n)}\]
\[\lim n \rightarrow \infty [1/2n(2n+3)\]
zero
This just tell us that it converge but not to what
yup,convergent
1/e= e^-1[1+1/2! +1/3!...]^-1 that can be expanded using binomial expension
\[[1+(1/2/!+1/3!...)]^{-1}=1- (1/2!+1/3!...)+(1/2!+ 1/3!...)^2....\]
how come the sum is 2/3!+4/5!+6/7!+....?
i dont kno. thats what the book says.. its wrong isn't it?
seems to be, not sure
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wow..great thinkin there buddy.. thanx!

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