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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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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.

  2. anonymous
    • 5 years ago
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    the series notation is wrong

  3. anonymous
    • 5 years ago
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    i dont think so buddy

  4. anonymous
    • 5 years ago
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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

  5. anonymous
    • 5 years ago
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    m with you..;-)

  6. anonymous
    • 5 years ago
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    1/2 is the answer?

  7. anonymous
    • 5 years ago
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    no its 1/e

  8. anonymous
    • 5 years ago
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    which book r u using?

  9. anonymous
    • 5 years ago
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    "rudiments of mathematics" its a book for wbjee(west bengal joint entnc xm)

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

  11. anonymous
    • 5 years ago
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    good!

  12. anonymous
    • 5 years ago
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    (2n+2)/(2(n+3) *(2n+1)!/2n

  13. anonymous
    • 5 years ago
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    (2n+2)/(2n+3)! *(2n+1)!/2n

  14. anonymous
    • 5 years ago
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    wow so u got the expression right finally

  15. anonymous
    • 5 years ago
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    \[2n+2 \over {(2n+2)(2n+3)(2n)}\]

  16. anonymous
    • 5 years ago
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    \[\lim n \rightarrow \infty [1/2n(2n+3)\]

  17. anonymous
    • 5 years ago
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    zero

  18. anonymous
    • 5 years ago
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    This just tell us that it converge but not to what

  19. anonymous
    • 5 years ago
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    yup,convergent

  20. anonymous
    • 5 years ago
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    1/e= e^-1[1+1/2! +1/3!...]^-1 that can be expanded using binomial expension

  21. anonymous
    • 5 years ago
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    \[[1+(1/2/!+1/3!...)]^{-1}=1- (1/2!+1/3!...)+(1/2!+ 1/3!...)^2....\]

  22. anonymous
    • 5 years ago
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    how come the sum is 2/3!+4/5!+6/7!+....?

  23. anonymous
    • 5 years ago
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    i dont kno. thats what the book says.. its wrong isn't it?

  24. anonymous
    • 5 years ago
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    seems to be, not sure

  25. nikvist
    • 5 years ago
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  26. anonymous
    • 5 years ago
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    wow..great thinkin there buddy.. thanx!

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