Resolved: "Does the sum ∑n=1∞n!(3+1)(3+2)…(3+n) converge or diverge?"

Adrienne Rowe

Answered question

2022-02-25

Does the sum

\sum_{n = 1}^{\infty} \frac{\sqrt{n!}}{(3 + \sqrt{1}) (3 + \sqrt{2}) \dots (3 + \sqrt{n})}

converge or diverge?

Lana Barlow

Beginner2022-02-26Added 4 answers

Let

a_{n} = \frac{\sqrt{n!}}{(3 + \sqrt{1}) (3 + \sqrt{2}) \dots (3 + \sqrt{n})} = \frac{1}{(1 + \frac{3}{\sqrt{1}}) (1 + 3 \sqrt{2}) \dots (1 + \frac{3}{\sqrt{n}})}

Then

\log (a_{n}) = - \sum_{k = 1}^{n} \log (1 + \frac{3}{\sqrt{k}})

In the sum, we have

\log (1 + \frac{3}{\sqrt{k}}) \sim \frac{3}{\sqrt{k}}

k \to \infty

, so

\log (a_{n}) \sim - \sum_{k = 1}^{n} \frac{3}{\sqrt{k}} \sim - 6 \sqrt{n}

n \to \infty

.
Now, it follows that

\frac{\log (a_{n})}{\log n} \to - \infty

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.

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