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I just can't seem to get my head around this sum:

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

The problem I'm having is how to solve the denominator. I tried d'Alembert's ratio test with no success.

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    Is it $\sqrt{(n!)}$ or $(\sqrt n)!$? – 2017-01-17
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    Also, can you show us what you did already? What happened when you applied that test? – 2017-01-17
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    It's the root of n!, with the factorial under the root. – 2017-01-17
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    I tried solving the denominator, and got (sqrt(n(n+1)/2))+.........+2. After applying d'Alembert: ((sqrt(n+1)!) / (sqrt((n+1)(n+2)/2)))+.........+2)) divided by (sqrt(n!) / (sqrt(n(n+1)/2))+.........+2). I'm quite new to this page so excuse my formatting! – 2017-01-17
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2 Answers 2

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A solution, maybe not the most elegant.

You have $$ a_n\stackrel{\rm def}{=}\frac{\sqrt{n!}}{\prod_{k=1}^n (2+\sqrt{k})} = \prod_{k=1}^n \frac{\sqrt{k}}{2+\sqrt{k}}. $$ Now, $$ \ln a_n = \sum_{k=1}^n \ln \frac{\sqrt{k}}{2+\sqrt{k}}= \sum_{k=1}^n \ln \frac{1}{1+\frac{2}{\sqrt{k}}}. $$ Since, when $k\to\infty$, $\ln \frac{1}{1+\frac{2}{\sqrt{k}}} =\ln \left(1-\frac{2}{\sqrt{k}}+o\left(\frac{2}{\sqrt{k}}\right)\right)=-\frac{2}{\sqrt{k}}+o\left(\frac{2}{\sqrt{k}}\right)$ and moreover the series $\sum_{k=1}^\infty \frac{2}{\sqrt{k}}$ diverges, we have by theorems of comparison that, when $n\to\infty$, $$ \ln a_n \operatorname*{\sim}_{n\to\infty} -\sum_{k=1}^n \frac{2}{\sqrt{k}} \operatorname*{\sim}_{n\to\infty} -4\sqrt{n} $$ from which $$ a_n = e^{-4\sqrt{n} + o(\sqrt{n})} \tag{$\ast$} $$ Now, by comparison (with e.g. $b_n\stackrel{\rm def}{=} e^{-n^{0.49}}$, which by $(\ast)$ satisfies $a_n = o(b_n)$ and $\sum_{n=1}^{\infty} b_n < \infty$), we get that the series $\sum_{n} a_n$ converges.

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    (+1) I was going along the same lines, you saved me from the trouble :) – 2017-01-17
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    My pleasure (this is the best way to procrastinate I found). – 2017-01-17
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Assuming $n\geq 4$, the general term is given by $$ a_n=\prod_{k=1}^{n}\frac{\sqrt{k}}{2+\sqrt{k}} = \prod_{k=1}^{n}\left(1+\frac{2}{\sqrt{k}}\right)^{-1}\leq\frac{1}{15}\prod_{k=4}^{n}\left(1+\frac{2}{\sqrt{k}}\right)^{-1}$$ and assuming $n\geq 10$ $$ \prod_{k=4}^{n}\left(1+\frac{2}{\sqrt{k}}\right)\geq 1+\!\!\!\!\!\sum_{\substack{A\subseteq\{\lfloor n/2\rfloor-1,\ldots,n\}\\|A|=4}}\prod_{a\in A}\frac{2}{\sqrt{a}}\geq\binom{n/2}{4}\frac{16}{n^2}\geq\frac{1}{24}\left(\frac{n}{2}-3\right)^2 $$ so the given series is convergent by comparison with $\sum_{n\geq 1}\frac{1}{n^2}$, that is convergent.