anonymous
  • anonymous
2/3!+4/5!+6/7!+....to infinity is equal to?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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.
anonymous
  • anonymous
the series notation is wrong
anonymous
  • anonymous
i dont think so buddy

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

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