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

- anonymous

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

- chestercat

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- Owlfred

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- watchmath

Please rewrite your series

- 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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## More answers

- watchmath

Press the equation button right below the box where you typed your question

- watchmath

I think we can use the ratio test here. Have you tried that?

- anonymous

yes i have tried that and i think this serie is diverge but i am not sure about that

- watchmath

It is absolutely convergent since the limit of \(|a_{n+1}|/|a_n|\) is 1/e.

- anonymous

\[\sum_{n=1}^{\infty} n!/(-n)^n\]

- anonymous

i think the right style is that

- 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

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

thank you very much . i understood it .

- watchmath

Great :)

- anonymous

\[an=\sqrt{2+\sqrt{2+\sqrt{...+\sqrt{2}}}}\]
is this converge?

- 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

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

and then what is the answer of the limit of an?

- watchmath

L=2 because -1 can't be the limit since the sequence is positive

- anonymous

thank you . that was a realy good answer.

- anonymous

\[\sum_{n=0}^{\infty} n.2^n+1/3^n+1 \]
is it converge?

- watchmath

of course not!! the 1 alone will make the series diverges

- anonymous

\[\sum_{n=1}^{\infty} n.r^n if 0

- watchmath

Post as a new question so others can help you. I am off to bed now :D.

- anonymous

sorry take a rest

- watchmath

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

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