I am not sure whether my argument works. I think I'm mostly concerned with how I'm working with the binomial coefficient. So I am wondering if I can get feed back as to whether my argument is correct/ the series is divergent or not.
Evaluate the following series:
$$\sum_{k=2}^{\infty} {3k \choose k} \cdot 7!$$
Let us begin by considering the sequence of partial sums:
$$ S_k = {3k \choose k} \cdot 7!$$
for $ k \in \mathbb{N}: k >1$. This expression can be expanded and rewritten to give:
$$ S_k = {3k \choose k} \cdot 7! = \left( \frac {(3k)!}{k! \cdot (2k)!} \right) \cdot 7! = \left( \frac{(3k) \cdot (3k – 1) \cdot … \cdot (2k + 1)}{k!} \right) \cdot 7! $$
Now, in order to evaluate the convergence, or lack thereof, of the infinite series we will consider the limit of the ratio $ \frac {S_{k + 1}}{S_k}$. In order to do this, we must first find an expression for $S_{k + 1}$:
$$S_{k+1} = {3(k+1) \choose k+1} \cdot 7! = \left( \frac {(3k + 3)!}{(k + 1)! \cdot (2 k + 2)!} \right) \cdot 7! = \left( \frac{(3k + 3) \cdot (3k + 2) \cdot … \cdot (2k + 3)}{(k + 1)!} \right) \cdot 7!$$
Finally, denote the aforementioned limit of the ratio $ \frac {S_{k + 1}}{S_k}$ by $Q$ and observe that:
$$Q = \lim_{k\to\infty} \frac{S_{k + 1}}{S_k} = \lim_{k\to\infty} \left( \frac{3k + 3) \cdot (3k + 2) \cdot … \cdot (2k + 3)}{(k + 1)!} \cdot \frac{k!}{(3k) \cdot (3k – 1) \cdot … \cdot (2k + 1)} \right)$$
$$ = \lim_{k\to\infty} \left( \frac {(3k + 3) \cdot (3k + 2) \cdot (3k + 1)}{(k + 1) \cdot (2k + 2) \cdot (2k + 1)} \right) = \frac {27}{4}$$
We have found $Q = \frac{27}{4}$ and thus, clearly, have that $Q > 1$. It follows from the Quotient test for absolute convergence that the series $\sum_{k=2}^{\infty} {3k \choose k} \cdot 7!$ is divergent, or:
$$\sum_{k=2}^{\infty} {3k \choose k} \cdot 7! = \infty$$
Best Answer
This argument works, but here are the ways to make it simpler: