Determine the convergence or divergence of the sequence

[zak1989]

Homework Statement

Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges find its limit.

a_n = (1*3*5*...*(2n-1))/(2n)ⁿ

Homework Equations

lim n->infinity a_n = L

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 a_n = 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 :)

 

[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?

 

[pasmith]

Homework Statement

Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges find its limit.

a_n = (1*3*5*...*(2n-1))/(2n)ⁿ

Homework Equations

lim n->infinity a_n = L

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 a_n = 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 :)

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).

 

[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.

 

[pasmith]

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?

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

 

[Mark44]

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

For n sufficiently large, 0 < 1/2n < 3/2n < 5/2n < ... < (2n-1)/2n < 1.

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.

 

[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?

 

[hilbert2]

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

 

[pasmith]

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?

If \epsilon &gt; 0, does there exist a positive integer N such that 1/(2N) &lt; \epsilon?

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

 

[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 :)