# Math Help - Another Sequence

1. ## Another Sequence

I don't understand how to deal with sequences like following:

$a_n=\frac{1\cdot 3\cdot 5\cdot...\cdot (2n-1)}{(2n)^n}$

It's not clear what the first term would be, or whether it converges. I want to list terms to see what is happening. If I write $a_k=\frac{(2k-1)^k}{(2k)^k}$ it is clear that the denominator increases at a faster rate as $k->\infty$, but I still can't tell what the entire sequence is going to do.

I don't understand how to deal with sequences like following:

$a_n=\frac{1\cdot 3\cdot 5\cdot...\cdot (2n-1)}{(2n)^n}$

It's not clear what the first term would be
I don't see why not. Some texts use the convention that sequences start with n=0 and other that sequences start with n= 1 but here it should be clear that if n= 0, so that 2n-1= -, the product in the numerator could not start with 1. Obviously the first term has n= 1 and is $a_1= \frac{1}{2^1}= \frac{1}{2}$.

, or whether it converges. I want to list terms to see what is happening. If I write $a_k=\frac{(2k-1)^k}{(2k)^k}$ it is clear that the denominator increases at a faster rate as $k->\infty$, but I still can't tell what the entire sequence is going to do.
It's not that difficult to see that the terms in the first sequence are $\frac{1}{2}$, $\frac{1(3)}{4^2}= \frac{3}{16}$, $\frac{1(3)(5)}{6^3}= \frac{15}{216}= \frac{5}{72}$. I would guess that it is rapidly convergent to 0.

It is useful to know that while any positive multiple of n increases without bound, any power of n increases faster, n! increases faster than any power of n, and $n^n$ increases faster than any multiple of n!. Technically, we would say that each dominates the previous. In determining convergence of a sequence we really only need to look at the "most dominant" term. If the numerator has greater "dominance" than the denominotor, the sequence diverges. If the denominator has greater "dominance" then the sequence converges to 0. In this sequence the numerator is a variation on n! while the denominator is basically $n^n$. The denominator dominates so this sequence goes to 0.

The numerator and denominator of the second sequence, $\frac{(2k-1)^k}{(2k)^k}= \left(\frac{2k-1}{2k}\right)^k$, have the same dominance so it converges to a non-zero number. In fact, it should not be difficult to see that for very very large k that "-1" in the numerator is comparitively 0 (If I had 10 trillion dollars, I would be willing to give you one!) so for large k, it is indistinguishable from $\left(\frac{2k}{2k}\right)^k= 1^k= 1$. The limit is 1.

3. I presume the first three terms would be:

$a_1=\frac{1}{2^1}=\frac{1}{2}$

$a_2=\frac{1\cdot 3}{4^2}=\frac{3}{16}$

$a_3=\frac{1\cdot 3\cdot 5}{6^3}=\frac{15}{216}$

What exactly do you wish to know?
The sum of the series? or something else?

$a_1=\frac{1}{2^1}=\frac{1}{2}$
$a_2=\frac{1\cdot 3}{4^2}=\frac{3}{16}$
$a_3=\frac{1\cdot 3\cdot 5}{6^3}=\frac{15}{216}$