how do i work out whether the sum to infinity (n=6) n!/(2n)! converges or not???

Answer 1

Well, the value of n!/(2n!) is equal to 1 / ((n + 1)(n + 2) ... (2 n))
This value is positive, but it is obviously less than 1 / n² when n is greater than or equal to 2. Because the sum of 1 / n² from 6 to infinity converges, the sum of n!/(2n!) must converge to something smaller.

Answer 2

Try taking the limit of the ratio of a term and its previous term.
This is known as the ratio test, and goes as follows:

lim |a[n + 1]/a[n]|
n -> infinity

If the result of this value is equal to r, then:
a) if r > 1, the series diverges.
b) if r < 1, the series converges.
c) if r = 1, the series may converge or diverge (inconclusive).


lim | ( [(n + 1)! / (2(n + 1)!)] / [ n! / (2n)! ] |
n -> infinity

Making this complex fraction into a simple fraction, we get

lim | ( [(n + 1)! (2n)!] / [ n! (2(n + 1))! ] |
n -> infinity

lim | ( [(n + 1)! (2n)!] / [ n! (2n + 2)! ] ) |
n -> infinity

Now, use the recursive properties of the factorial. Note that
(n + 1)! = (n + 1)n!, and
(2n + 2)! = (2n + 2)(2n + 1)! = (2n + 2)(2n + 1)(2n)!

lim | ( [(n + 1)n! (2n)!] / [ n! (2n + 2)(2n + 1)(2n)! ] |
n -> infinity

Now, make the appropriate cancellations.

lim | ( [n + 1] / [(2n + 2)(2n + 1)] ) |
n -> infinity

We can using Calculus I concepts to know that this limit is equal to 0.

Therefore, the series converges.

Answer 3

I don't know what the "n=6" here means. I'll assume the series is ∑ n!/(2n)!

One way to show convergence is to compare the terms to that of a known series. Each term of this series is
n! / (2n)! =
1 / [2n * (2n - 1) * (2n - 2) * ... * (n+2)*(n+1) ] =
1 / [(2n² + 2n) * (2n - 1) * (2n - 2) * ... * (n+2) ]

So we can at least say that each term is less than 1/(2n²). The series ∑1/(2n²) is a geometric series so it converges. If each term of our series is smaller than that, then it must coverge too.

Answer 4

ratio test
a(sub n) = n!/(2n)!
then
a(sub n +1)/a(sub n)=
(n+1)!/(2(n+1))! * (2n)!/(n)!

and we know that (n+1)! = (n+1)(n)(n-1)...
and n! = (n)(n-1)...so
(n+1)! = (n+1)n!

= (n+1)(n)!/((2n+2)! * (2n)!/n!

the n! cancels
=(n+1)* (2n)!/((2n+2)(2n+1)(2n)!)
=(n+1)/((2n+2)(2n+1))
and the limit as n approaches infinity is 0 which is less than 1
so it converges by the ratio test

Answer 5

use the ratio test;

u(n)=n!/(2n)!
u(n+1)=(n+1)!/(2n+1)!
therefore,
u(n+1)/un
={(n+1)!/(2n+1)!}/{n!/(2n)!}
={(n+1)!/(2n+1)!}*{(2n)!/n!}
=(n+1)/(2n+1)
when n tends to infinity,
(n+1)/(2n+1) tends to 1/2

since,for all values of n,
the ratio u(n+1)/u(n)< than
some fixed number less
than 1,the series [sigma]u(n)
is convergent

i hope that this helps