I'm having difficulty understanding how the following converges to zero:
$\displaystyle a_n=\frac{1*3*5*...*(2n-1)}{(2n)^n}$
I've been looking through my book, and I can't tell what theorem I need to use.
I believe ratio-test is only useful for determining convergence of a sum $\displaystyle \sum_{n=1}^{\infty}a_n$
You can show however $\displaystyle a_{n+1}< a_{n} $
wich is: $\displaystyle \frac{1\cdot 3\cdots (2n+1)}{(2n+2)^{n+1}}< \frac{1\cdot 3\cdots (2n-1)}{(2n)^n}\Leftrightarrow \frac{2n+1}{(2n+2)^{n+1}}< \frac{1}{(2n)^n}$
This last inequality is easy to proof.
If the sum converges, then the sequence must converge to 0!
But that doesn't prove it converges to 0.You can show however $\displaystyle a_{n+1}< a_{n} $
wich is: $\displaystyle \frac{1\cdot 3\cdots (2n+1)}{(2n+2)^{n+1}}< \frac{1\cdot 3\cdots (2n-1)}{(2n)^n}\Leftrightarrow \frac{2n+1}{(2n+2)^{n+1}}< \frac{1}{(2n)^n}$
This last inequality is easy to proof.
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.