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anonymous
 5 years ago
Is this ∑_(n=1)^∞▒〖n!÷(n)^n〗converge?
anonymous
 5 years ago
Is this ∑_(n=1)^∞▒〖n!÷(n)^n〗converge?

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! 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!

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Please rewrite your series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0∑_(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.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Press the equation button right below the box where you typed your question

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1I think we can use the ratio test here. Have you tried that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i have tried that and i think this serie is diverge but i am not sure about that

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1It is absolutely convergent since the limit of \(a_{n+1}/a_n\) is 1/e.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} n!/(n)^n\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think the right style is that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you for your answer .can you write your answer completely for me because i really want to know how you solve it.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0thank you very much . i understood it .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[an=\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}\] is this converge?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Yes, 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^2L+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
 5 years ago
Best ResponseYou've already chosen the best response.1L^2L+2=0 (L2)(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
 5 years ago
Best ResponseYou've already chosen the best response.0and then what is the answer of the limit of an?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1L=2 because 1 can't be the limit since the sequence is positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you . that was a realy good answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=0}^{\infty} n.2^n+1/3^n+1 \] is it converge?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1of course not!! the 1 alone will make the series diverges

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} n.r^n if 0<r<1\] what about that?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Post as a new question so others can help you. I am off to bed now :D.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1You can also ask me here: http://ask.watchmath.com when I am not around. See you!
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