I was given the following equation and told to prove either convergence or divergence. I am sure that it diverges, but I am unsure as to how I would go about proving that mathematically.
$$ \sum_{k=0}^n \frac{(-2^k)*1*3*5...(2k+1)}{1*4*7...(3k+1)}\ $$
I used the ratio test as I remember I was supposed to, but I cannot figure out how to cancel the remaining terms out to get a usable answer. Does it leave me with just:
$ \frac{(2)(2k+3)(3k+10)}{(3k+4)(2k+1)}\ $
That seems wrong to me but it's as far as I could go
You might have a typo, probably the question has $(-2)^k$ rather than $-2^k$. But it does not affect the analysis. If a series $\sum_{k=0}^n a_k$ is convergent, it is necessarily true that $\lim_{k\rightarrow \infty} a_k = 0$. Note that $$ a_k = \frac{(-2)^k*1*3*5...(2k+1)}{1*4*7...(3k+1)}. $$ The number of terms in the numerator and the denominator are equal to $k+1$ (except $(-2)^k$). $$ a_k = (-1)^n\frac{1}{2}\frac{2 \times 1}{1}\frac{2 \times 3}{4}\cdots \frac{2 \times (2k+1)}{(3k+1)}. $$ You can see that the right hand side terms are each greater than 1 except a constant. Moreover, the last term goes to infinity, which implies that the term $a_k$ have two subsequences going to $+\infty$ and $-\infty$. Therefore, this series cannot be convergent.