1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

If we take three or four, then the test series will still be convergent.

Sure, those would be alternate valid ways of proving the convergence of your original series.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .