# Thread: Convergence of a sequence.

1. ## Convergence of a sequence.

I'm having difficulty understanding how the following converges to zero:

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

I've been looking through my book, and I can't tell what theorem I need to use.

2. I would use the ratio test. Write out the exponentiation in the denominator, and compare it with the product in the numerator.

3. I believe ratio-test is only useful for determining convergence of a sum $\sum_{n=1}^{\infty}a_n$

You can show however $a_{n+1}< a_{n}$

wich is: $\frac{1\cdot 3\cdots (2n+1)}{(2n+2)^{n+1}}< \frac{1\cdot 3\cdots (2n-1)}{(2n)^n}\Leftrightarrow \frac{2n+1}{(2n+2)^{n+1}}< \frac{1}{(2n)^n}$

This last inequality is easy to proof.

4. Originally Posted by Dinkydoe
I believe ratio-test is only useful for determining convergence of a sum $\sum_{n=1}^{\infty}a_n$
If the sum converges, then the sequence must converge to 0!

You can show however $a_{n+1}< a_{n}$

wich is: $\frac{1\cdot 3\cdots (2n+1)}{(2n+2)^{n+1}}< \frac{1\cdot 3\cdots (2n-1)}{(2n)^n}\Leftrightarrow \frac{2n+1}{(2n+2)^{n+1}}< \frac{1}{(2n)^n}$

This last inequality is easy to proof.
But that doesn't prove it converges to 0.

5. I just realized why ratio-test does work, wanted to correct my statement.

...

forget it, I'm convinced. Ratio-test must be used here.

6. Originally Posted by HallsofIvy
If the sum converges, then the sequence must converge to 0!

But that doesn't prove it converges to 0.
The solution to the problem I posted can be found here:

Calc Chat Free Solutions

Calculus, 9th Edition, Chapter 9, Section 1, Excercise 53

I don't understand the solution, or how they proved it was zero. In section 9.1 of my book, the only theorems discussed are The Squeeze Theorem, The Absolute Value Theorem, and the Bounded Monotonic Sequence Theorem.