TITLE: Does the sum $\sum_{n = 1} ^\infty \frac{\sqrt {n!}} {(3+\sqrt 1)(3+\sqrt 2)...(3+\sqrt n)} $ converge or diverge?
QUESTION [8 upvotes]: The task is to study the convergence of the series $\sum_{n = 1} ^\infty \frac{\sqrt {n!} } {(3+\sqrt 1)(3+\sqrt 2)...(3+\sqrt n)} $. I've tried applying Gauss's and ratio test but it didn't get me anywhere. In my textbook it says that this series is convergent.

REPLY [8 votes]: Let $$a_n := \frac{\sqrt{n!}}{(3+\sqrt{1})(3+\sqrt{2})\cdots(3+\sqrt{n})} = \frac{1}{(1+3/\sqrt{1})(1+3/\sqrt{2})\cdots(1+3/\sqrt{n})}.$$
Then
$$\log(a_n) = -\sum_{k=1}^n \log\left(1 + \frac{3}{\sqrt{k}}\right).$$
In this sum, we have $\log(1 + 3/\sqrt{k}) \sim \frac{3}{\sqrt{k}}$ as $k \to \infty$, so $\log(a_n) \sim -\sum_{k=1}^n \frac{3}{\sqrt{k}} \sim -6 \sqrt{n}$ as $n \to \infty$.
Now, it follows that $\frac{\log(a_n)}{\log n} \to -\infty$ as $n \to \infty$, so for $n$ sufficiently large, we have $\frac{\log(a_n)}{\log n} < -2$.  Therefore, for $n$ sufficiently large, $a_n < \frac{1}{n^2}$, and by the comparison test we can conclude that $\sum_{n=1}^\infty a_n$ converges.