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I have a series with the $n$-th member defined by the following equation

$$a_n = \frac{n!e^n}{n^{n+\alpha}}, \alpha \in \mathbb{R}$$

The series starts at $n=1$.

Questions

  1. Is it possible to figure out whether the series converges using the ratio test? Because it works out to 1 for me.
  2. How to approach factorials when dealing with series
  3. What method would be the most appropriate here to test for convergence?
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Morgan Wilde
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2 Answers2

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The best method here is to use the Stirling formula: $$n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$$

Thus $$a_n \sim \sqrt{2\pi}n^{1/2-\alpha}$$

Using the fact that if two positive sequences ( it suffices to you know that one of the two is positive) are such that $a_n \sim b_n$ then $\sum a_n$ and $\sum b_n$ have the same nature in terms of convergence, it follows that

$\sum a_n$ converges iff $\alpha - \frac{1}{2}>1$ ie $\alpha > 3/2$

Gabriel Romon
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  1. Yes, the ratio for the ratio test converges to $1$. That is when we say "the ratio test fails". It does not tell you whether you have convergence or not.

  2. Factorials are great for the ratio test, since they mostly cancel. In this test, it was easy to do the limit. Failure of the test is another question.

  3. In this case, if you can do some asymptotics, $$ \frac{a_n}{a_{n+1}} = \frac{1}{e}\left(1+\frac{1}{n}\right)^{n+\alpha} = 1 + \frac{\alpha-\frac{1}{2}}{n} + O\left(\frac{1}{n^2}\right) \qquad\text{as } n \to \infty $$ so by Gauss's Test we see that the series $\sum a_n$ converges if $\alpha >\frac{3}{2}$ and diverges if $\alpha \le \frac{3}{2}$

GEdgar
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