# Series and convergence

1. Nov 13, 2005

### thenewbosco

to solve this question i need to know whether the following series converges. however both the root test and ratio test give 1.

sum from n=1 to infinity:
$$\sum\frac{-1(2n-2)!}{n!(n-1)!2^{2n-1}}$$

Last edited: Nov 13, 2005
2. Nov 13, 2005

### benorin

Do you mean $$\sum\frac{(-1)^{n}(2n-2)!}{n!(n-1)!2^{2n-1}}$$ or $$-\sum\frac{(2n-2)!}{n!(n-1)!2^{2n-1}}$$ ?

3. Nov 13, 2005

### thenewbosco

yes, the second one as you have written it, with the -1 out front

4. Nov 13, 2005

### benorin

5. Nov 13, 2005

### thenewbosco

this doesnt seem to help me in any way..

6. Nov 13, 2005

### benorin

Does $$-\sum_{n=1}^{\infty}\frac{(2n-2)!}{n!(n-1)!2^{2n-1}}=-1$$ help?

7. Nov 13, 2005

### benorin

Try partial sums

Try partial sums:

Prove that:

$$S_{N}:=-\sum_{n=1}^{N}\frac{(2n-2)!}{n!(n-1)!2^{2n-1}}= \frac{\left( 2N\right)!}{2^{2N}\left( N!\right)^2} -1 = \frac{1}{2^{2N}} \left(\begin{array}{cc}2N\\N\end{array}\right)-1$$

by induction (I hope induction will work, anyway) where $$\left(\begin{array}{cc}n\\k\end{array}\right)$$ is a binomial coefficient, and then show that the partial sum $$S_{N} \rightarrow 1\mbox{ as } N\rightarrow \infty$$.

But I won't kid you, this is NOT the easy way to do this problem.