anonymous
  • anonymous
Is this ∑_(n=1)^∞▒〖n!÷(-n)^n〗converge?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Owlfred
  • Owlfred
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watchmath
  • watchmath
Please rewrite your series
anonymous
  • anonymous
∑_(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.

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watchmath
  • watchmath
Press the equation button right below the box where you typed your question
watchmath
  • watchmath
I think we can use the ratio test here. Have you tried that?
anonymous
  • anonymous
yes i have tried that and i think this serie is diverge but i am not sure about that
watchmath
  • watchmath
It is absolutely convergent since the limit of \(|a_{n+1}|/|a_n|\) is 1/e.
anonymous
  • anonymous
\[\sum_{n=1}^{\infty} n!/(-n)^n\]
anonymous
  • anonymous
i think the right style is that
anonymous
  • anonymous
thank you for your answer .can you write your answer completely for me because i really want to know how you solve it.
watchmath
  • watchmath
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 \[\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n=1/e\]
anonymous
  • anonymous
thank you very much . i understood it .
watchmath
  • watchmath
Great :)
anonymous
  • anonymous
\[an=\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}\] is this converge?
watchmath
  • watchmath
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.
watchmath
  • watchmath
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.
anonymous
  • anonymous
and then what is the answer of the limit of an?
watchmath
  • watchmath
L=2 because -1 can't be the limit since the sequence is positive
anonymous
  • anonymous
thank you . that was a realy good answer.
anonymous
  • anonymous
\[\sum_{n=0}^{\infty} n.2^n+1/3^n+1 \] is it converge?
watchmath
  • watchmath
of course not!! the 1 alone will make the series diverges
anonymous
  • anonymous
\[\sum_{n=1}^{\infty} n.r^n if 0
watchmath
  • watchmath
Post as a new question so others can help you. I am off to bed now :D.
anonymous
  • anonymous
sorry take a rest
watchmath
  • watchmath
You can also ask me here: http://ask.watchmath.com when I am not around. See you!

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