1. ## Converge or Diverge?

$\displaystyle a_k=\frac{4^k(k!)^2}{(2k)!}$

Does $\displaystyle \sum_1^\infty (-1)^ka_k$ converge absolutely, converge conditionally, or diverge?

Any help on this? Not really sure where to even begin...

2. Originally Posted by paupsers
$\displaystyle a_k=\frac{4^k(k!)^2}{(2k)!}$

Does $\displaystyle \sum_1^\infty (-1)^ka_k$ converge absolutely, converge conditionally, or diverge?

Any help on this? Not really sure where to even begin...
absolute convergence means the series $\displaystyle \sum |a_k|$ converges. Conditional convergence means $\displaystyle \sum (-1)^ka_k$ converges, but $\displaystyle \sum a_k$ doesn't.

Here, I would use the ratio test. so try that.

3. if $\displaystyle \sum a_n<\infty,$ and obviously here $\displaystyle a_n\ge0,$ then $\displaystyle \sum(-1)^n a_n<\infty,$ and this is obvious since $\displaystyle \sum |(-1)^n a_n|=\sum a_n<\infty.$

4. Originally Posted by paupsers
$\displaystyle a_k=\frac{4^k(k!)^2}{(2k)!}$

Does $\displaystyle \sum_1^\infty (-1)^ka_k$ converge absolutely, converge conditionally, or diverge?

Any help on this? Not really sure where to even begin...
I can tell you without computation that $\displaystyle a_k\geq 1$ for all $\displaystyle k$, which pretty much settles it...

So, why? Because $\displaystyle \frac{1}{a_k}=\frac{1}{2^{2k}}{2k\choose k}$ is the probability for a symmetric simple random walk on the integers to be back at its starting point at time $\displaystyle 2k$ (the binomial coefficient comes from choosing which $\displaystyle k$ steps go to the right); hence $\displaystyle \frac{1}{a_k}\leq 1$. Or just because $\displaystyle {2k\choose k}$ is the number of subsets of $\displaystyle \{1,\ldots, 2k\}$ with $\displaystyle k$ elements, which is less than $\displaystyle 2^{2k}$, the number of subsets of $\displaystyle \{1,\ldots,2k\}$.

By the way, Stirling estimate gives $\displaystyle a_n\sim\sqrt{\pi n}$ if I remember well, hence the ratio and root tests would be unconclusive.