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anonymous

  • 5 years ago

Is this ∑_(n=1)^∞▒〖n!÷(-n)^n〗converge?

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. watchmath
    • 5 years ago
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    Please rewrite your series

  3. anonymous
    • 5 years ago
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    ∑_(n=1)^∞〖n!/(-n)^n〗i wrote it on the office word and when i copy it on the web page it changed like this. please help me . I don't know where i have to type my series and the formula.

  4. watchmath
    • 5 years ago
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    Press the equation button right below the box where you typed your question

  5. watchmath
    • 5 years ago
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    I think we can use the ratio test here. Have you tried that?

  6. anonymous
    • 5 years ago
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    yes i have tried that and i think this serie is diverge but i am not sure about that

  7. watchmath
    • 5 years ago
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    It is absolutely convergent since the limit of \(|a_{n+1}|/|a_n|\) is 1/e.

  8. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{\infty} n!/(-n)^n\]

  9. anonymous
    • 5 years ago
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    i think the right style is that

  10. anonymous
    • 5 years ago
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    thank you for your answer .can you write your answer completely for me because i really want to know how you solve it.

  11. watchmath
    • 5 years ago
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    Yes, now\[|a_{n+1}/a_n|=|\frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^n}{n!}|=\frac{n!(n+1)}{(n+1)\cdot (n+1)^n}\cdot \frac{n^n}{n!}=\left(\frac{n}{n+1}\right)^n\] Now remember that \[\lim_{n\to\infty} \left(\frac{n+1}{n}\right)^n=\lim(1+\frac{1}{n})^n=e\] It follows \[\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n=1/e\]

  12. anonymous
    • 5 years ago
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    thank you very much . i understood it .

  13. watchmath
    • 5 years ago
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    Great :)

  14. anonymous
    • 5 years ago
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    \[an=\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}\] is this converge?

  15. watchmath
    • 5 years ago
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    Yes, you need to show first that the limit exist by showing that the sequence is increasing and bounded above. Once we already establish that computing the limit is not that difficult. Suppose the limit is L, then L^2=2+L. So the limit is the root of L^2-L+2=0. Knowing this root can help us also to find the right upperbound when we are trying to show that the sequence is bounded.

  16. watchmath
    • 5 years ago
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    L^2-L+2=0 (L-2)(L+1)=0 L=2 or L=-1 So it is reasonable to make a conjecture that an is bounded above by 2 and try to prove that by induction.

  17. anonymous
    • 5 years ago
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    and then what is the answer of the limit of an?

  18. watchmath
    • 5 years ago
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    L=2 because -1 can't be the limit since the sequence is positive

  19. anonymous
    • 5 years ago
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    thank you . that was a realy good answer.

  20. anonymous
    • 5 years ago
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    \[\sum_{n=0}^{\infty} n.2^n+1/3^n+1 \] is it converge?

  21. watchmath
    • 5 years ago
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    of course not!! the 1 alone will make the series diverges

  22. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{\infty} n.r^n if 0<r<1\] what about that?

  23. watchmath
    • 5 years ago
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    Post as a new question so others can help you. I am off to bed now :D.

  24. anonymous
    • 5 years ago
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    sorry take a rest

  25. watchmath
    • 5 years ago
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    You can also ask me here: http://ask.watchmath.com when I am not around. See you!

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