# Determine the convergence or divergence of the sequence

1. Jul 31, 2013

### zak1989

1. The problem statement, all variables and given/known data
Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges find its limit.

an = (1*3*5*....*(2n-1))/(2n)n

2. Relevant equations
lim n->infinity an = L

3. The attempt at a solution
The answer in the book shows:
1/2n * 3/2n * 5/2n....(2n-1)/2n less than 1/2n

So, lim n->infinity an = 0, converges

I don't understand where the 1/2n * 3/2n * 5/2n....(2n-1)/2n less than 1/2n comes from. I'm obviously missing something. Could anyone please explain that part to me? I'd really appreciate any help at all. Thanks :)

2. Jul 31, 2013

### hilbert2

Obviously $1 \cdot 2 \cdot 3 \dots (2n-1) \cdot (2n) < (2n)^{n}$, because in the first product all factors except the last one are smaller than in the second product.

Therefore, also $\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{(2n)^{n}}<\frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{1 \cdot 2 \cdot 3 \dots (2n-1) \cdot (2n)}=\frac{1}{2 \cdot 4 \dots (2n)}$.

Can you now tell that the sequence is convergent?

3. Jul 31, 2013

### pasmith

For n sufficiently large, 0 < 1/2n < 3/2n < 5/2n < ... < (2n-1)/2n < 1. So you're multiplying 1/(2n) by something which is strictly less than 1, which means that the result is strictly less than 1/(2n).

4. Jul 31, 2013

### zak1989

Thanks for your help, but I'm still not getting it. When I see the sequence an = (1*3*5*....*(2n-1))/(2n)^n I would think it looks something like 1/2(1)^1 * 3/2(2)^2 * 5/2(3)^3 .... Is that wrong? I don't understand where the exponent n is going from the denominator.

5. Jul 31, 2013

### pasmith

Yes; it's
$$\frac1{2n} \times \frac3{2n} \times \cdots \times \frac{2n-1}{2n} = \frac{1 \times 3 \times \cdots \times (2n-1)}{(2n)^n}$$
since there are $n$ factors and the denominator is the same for each.

6. Jul 31, 2013

### Staff: Mentor

Both, if you mean $\frac 1 {2n}$, $\frac 3 {2n}$, and so on, you need parentheses around each denominator. Otherwise what both of you wrote would be interpreted as (1/2) * n, (3/2) * n, etc.

Last edited: Jul 31, 2013
7. Jul 31, 2013

### zak1989

Ok now I understand why those two equations are equal, thanks. It seems so obvious now. What I still don't understand is how does saying 1/2n * 3/2n * 5/2n....(2n-1)/2n less than 1/2n help me conclude that the sequence diverges?

8. Jul 31, 2013

### hilbert2

^ Actually, the sequence converges, because you can make the quotient arbitrarily close to zero by choosing sufficiently large n.

9. Jul 31, 2013

### pasmith

If $\epsilon > 0$, does there exist a positive integer $N$ such that $1/(2N) < \epsilon$?

What does that tell you about the convergence or divergence of the sequence?

10. Jul 31, 2013

### zak1989

Oh ok so the limit as n goes to infinity is zero which means it converges, but I'm a little lost. How do I know the limit is zero? Thanks for all your help I really appreciate it :)