Series and convergence of a factorial

[Guest]

Hi can someone help me with the fatorial and prove why is it zero? I don't understand it and thus I can't apply it to other factorials in my calculus problems to see if it's divergent or convergent such as 1/(n+1)!

\(\displaystyle \sum_{n=1}^\infty \frac{1}{n!}\)

Thanks.

And also, I don't quite understand this. Can someone please explain it to me word for word that is understandable, preferiably in laymen's term. Thanks.

 

[galactus]

You can use the ratio test.

\(\displaystyle \lim_{n\to\infty}\frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}\)

=\(\displaystyle \lim_{n\to\infty}\frac{n!}{(n+1)!}\)

=\(\displaystyle \lim_{n\to\infty}\frac{1}{n+1}=0<1\)

Therefore, the series converges. But do you know what it convergses to?.

Write out the series:

\(\displaystyle \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{n!}\)

Compare this to the series for e.




Write it down and cancel. It becomes obvious:

\(\displaystyle \frac{(2n-1)(2n-2)(2n-3)....(2n-k)}{(2n+1)(2n)(2n-1)(2n-2)(2n-3)....(2n-k)}\)

As you can see, they all cancel except for the 2n and 2n+1 and you have:

\(\displaystyle \frac{1}{2n(2n+1)}\)

Does this help?.

 

[Guest]

How did you get the 2n?

 

[pka]

Do you understand that: \(\displaystyle 9! = \left( 9 \right)\left( 8 \right)\left( 7 \right)\left( 6 \right)\left( 5 \right)\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)\)
and that \(\displaystyle 10! = (10)\left( 9 \right)\left( 8 \right)\left( 7 \right)\left( 6 \right)\left( 5 \right)\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right) = (10)(9!)\)?

Now it follows that \(\displaystyle \left[ {2\left( 5 \right)} \right]! = (10!) = (10)(9!) = (10)(9)(8!)\).

So that \(\displaystyle \frac{{\left[ {2\left( 5 \right) - 1} \right]!}}{{\left[ {2\left( 5 \right) + 1} \right]!}} = \frac{{9!}}{{11!}} = \frac{{9!}}{{(11)(10)(9!)}} = \frac{1}{{(11)(10)}}\).

 

[galactus]

Check out my post. Because (2n+1)! is

(2n+1)(2n)(2n-1)(2n-2)(2n-3)......(2n-k).

Correct?. That's a factorial. The product of the first n positive integers.

Enter in a value for n if it makes it a little easier.

(2(2)+1)(2(2))(2(2)-1)(2(2)-2)(2(2)-3)=5*4*3*2*1