Question

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

Answer 1

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!

Answer 2

Please rewrite your series

Answer 3

∑_(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.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

i think the right style is that

Answer 10

thank you for your answer .can you write your answer completely for me because i really want to know how you solve it.

Answer 11

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 \[\lim_{n\to\infty} \left(\frac{n}{n+1}\right)^n=1/e\]

Answer 12

thank you very much . i understood it .

Answer 13

Great :)

Answer 14

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

Answer 15

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.

Answer 16

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.

Answer 17

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

Answer 18

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

Answer 19

thank you . that was a realy good answer.

Answer 20

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

Answer 21

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

Answer 22

\[\sum_{n=1}^{\infty} n.r^n if 0<r<1\]
what about that?

Answer 23

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

Answer 24

sorry take a rest

Answer 25

You can also ask me here:
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when I am not around.
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