# Direct Comparison Test: Choosing a Test Series

Determine if $$\sum\limits_{i = 1}^{\infty} \frac{i!}{i^i}$$ converges using the Direct Comparison Test.

Answer: Since $$\begin{eqnarray*} \frac{i!}{i^i} & = & \frac{1\cdot 2\cdot 3\cdots i}{i\cdot i\cdot i\cdots i} \\ \frac{i!}{i^i} & = & \frac{1}{i}\cdot\frac{2}{i} \cdot\frac{3}{i} \cdots \frac{i}{i} \\ \frac{i!}{i^i} & \leq & \frac{1}{i}\cdot\frac{2}{i} \\ \frac{i!}{i^i} & \leq & \frac{2}{i^2} \\ a_i & \leq & b_i \end{eqnarray*}$$

Since $$\sum\limits_{i = 1}^{\infty} \frac{2}{i^2}$$ is a convergent $$p$$-series and $$a_i \leq b_i$$, then the given series converges by the Direct Comparison Test.

However, my concern is why in the approximation are only the terms $$\frac{1}{i}\cdot\frac{2}{i}$$ selected? If we take three or four, then the test series will still be convergent. If we take only one, the test series will be divergent.

Right. This leads to a vacuous upper bound $$\sum_i \frac{i!}{i^i} \le \sum_i \frac{1}{i} = \infty$$ and does not provide any information.