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watchmath

  • 5 years ago

Interesting problem anyone? Compute \[\lim_{n\to\infty}\frac{1\cdot 1!+2\cdot 2!+\cdots+n\cdot n!}{(n+1)!}\]

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  1. anonymous
    • 5 years ago
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    guess or proof?

  2. anonymous
    • 5 years ago
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    just kidding.

  3. watchmath
    • 5 years ago
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    guess is welcome too :D

  4. anonymous
    • 5 years ago
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    1

  5. watchmath
    • 5 years ago
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    Good guess!! :D

  6. anonymous
    • 5 years ago
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    think i even know what it is, but the proof is eluding me so now i have something to think about for the afternoon.

  7. anonymous
    • 5 years ago
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    closed form i mean.

  8. anonymous
    • 5 years ago
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    won't post until i prove it.

  9. watchmath
    • 5 years ago
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    alrighty!

  10. anonymous
    • 5 years ago
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    Can we not just factor out an n! from each term in the top, and rewrite it as: \[\lim_{n \rightarrow \infty} \frac{n\cdot n![\frac{1}{n\cdot n!} + \frac{2 \cdot 2!}{n\cdot n!} + ... + 1]}{n\cdot n!(1+\frac{1}{n})} = \lim_{n \rightarrow \infty} \frac{\frac{1}{n\cdot n!} + \frac{2 \cdot 2!}{n\cdot n!} + ... + 1}{(1+\frac{1}{n})} = \frac{1}{1} = 1\]

  11. anonymous
    • 5 years ago
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    I guess it's not clear that the sum of all those n terms will go to 0 faster than the number of terms are growing but it seems reasonable to me.

  12. watchmath
    • 5 years ago
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    Yes, I think it works. We just need to show more rigorously that \[\lim_{n\to\infty}\frac{(n-r)\cdot(n-r)!}{n\cdot n!}=0\] for all \(0<r<n\).

  13. anonymous
    • 5 years ago
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    yeah, I'm looking at that now actually ;)

  14. anonymous
    • 5 years ago
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    actually just showing it works for r=1 will show it for the rest

  15. watchmath
    • 5 years ago
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    Agree :).

  16. anonymous
    • 5 years ago
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    I found it

  17. anonymous
    • 5 years ago
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    \[(n-1)! = \frac{n!}{n} \rightarrow \lim_{n\to\infty}\frac{(n-1)\cdot (n-1)!}{n\cdot n!} = \lim_{n\to\infty}\frac{(n-1)\cdot n!}{n^2 n!} = \lim_{n\to\infty}\frac{(n-1)}{n^2} = 0\]

  18. anonymous
    • 5 years ago
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    i think it is just \[\frac{(n+1)!-1}{(n+1)!}\]yes?

  19. watchmath
    • 5 years ago
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    Awesome satellite! :D

  20. anonymous
    • 5 years ago
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    proof by induction as soon as i figure it out.

  21. anonymous
    • 5 years ago
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    My method doesn't work?

  22. watchmath
    • 5 years ago
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    Your method works! Satellite shows that the expression is in fact equal to \(\frac{(n+1)!-1}{(n+1)!}\)

  23. anonymous
    • 5 years ago
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    Oh I see

  24. anonymous
    • 5 years ago
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    well actually satellite has not shown anything. i just said it.

  25. anonymous
    • 5 years ago
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    so no credit yet that is for sure.

  26. anonymous
    • 5 years ago
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    Fun problem though

  27. anonymous
    • 5 years ago
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    i thought it would be simple induction. but i am getting stuck because i keep getting \[n(n+1)!+2(n+1)!\] and i need this to be \[(n+2)!\] which, if it is, is not clear to me.

  28. anonymous
    • 5 years ago
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    any hints?

  29. watchmath
    • 5 years ago
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    Are you sure you want a hint. I am afraid it will soil the fun :).

  30. anonymous
    • 5 years ago
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    Just factor

  31. anonymous
    • 5 years ago
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    ok no hint. perhaps induction is not way to go?

  32. anonymous
    • 5 years ago
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    oh lord i get the dumb guy award

  33. watchmath
    • 5 years ago
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    ok, half hint. The summation on the top can be made into a telescoping sum.

  34. anonymous
    • 5 years ago
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    hehe.. sometimes we can't see the forest for the trees. Happens to us all.

  35. anonymous
    • 5 years ago
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    polpak got it. proof by induction done.

  36. anonymous
    • 5 years ago
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    but now i have to think about the telescope.

  37. watchmath
    • 5 years ago
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    \((n+1)!-1=(n+1)!-1!\) :D

  38. watchmath
    • 5 years ago
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    I think a hint won't hurt now :D. \(n\cdot n!=((n+1)-1)\cdot n!\)

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