The task is to prove, that series: $$\sum_{n=2}^\infty \frac{1}{log^2(n!)}$$ converges. Unfortunately I only managed to deal with showing that Limit of the summand is equal to 0, what is pretty obvious, but still, I am truly clueless, how to show that it converges. I am not asking for solution, but only clues/hints and other stuff like that. I would be truly grateful for any help.
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Are you familiar with Stirling's formula? – Antonio Vargas Nov 21 '15 at 15:18
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Unfortunetly...no. – Kiwi Nov 21 '15 at 15:19
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2Note that eventually $n!>e^n$ . – Aitor Ormazabal Nov 21 '15 at 15:20
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3Show that for $n\geq 4$, you have $n!\geq 2^n$. – Kelenner Nov 21 '15 at 15:20
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@Aitor Ormazabal It's not like that to prove that one series converges by comparison test, the "bigger" one should converge, not the smaller one? – Kiwi Nov 21 '15 at 15:22
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Use $n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ And substitute into the log. argument, then use log properties for products... – Enrico M. Nov 21 '15 at 15:24
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@Henry: I suspect that Aitor Ormazabal and Kelenner and Thomas's observations are a simpler approach, depending on the base of the loagrithm – Henry Nov 21 '15 at 15:26
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1Yes, but taking reciprocals reverses the inequality: If you can prove that $log^2(n!)>n^2$ then you have $\dfrac {1}{log^2(n!)}<\dfrac {1}{n^2}$ and you can use comparison – Aitor Ormazabal Nov 21 '15 at 15:26
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My mistake, sorry. Also great idea. Thanks! – Kiwi Nov 21 '15 at 15:32
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That series cannot converge. Series starts from $n = 1$ which is a pole for the $\log$ functions. First term diverges yet! – Enrico M. Nov 21 '15 at 17:57
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Hint: you can easily show that $\ln(n!) \geq c n\ln n$ for some absolute constant $c >0$. From there, $\frac{1}{\ln^2 n!} \leq \frac{1}{c^2 n^2\ln^2 n}$, and conclude by theorems of comparison.
Clement C.
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Similar to the comments above, you first note that $n! > e^n$ for $n\geq k$ (for some $k$). Then you can say that for $n\geq k$ $$ \log(n!)^2 > (\log(e^n))^2 = (n\log(e))^2 = n^2. $$ So $$ \frac{1}{\log(n!)^2} < \frac{1}{n^2}. $$ Then you can use the comparison test.
Thomas
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The series cannot converge, it starts from $n = 1$: first terms diverges yet! – Enrico M. Nov 21 '15 at 17:58
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By summation by parts, for any $n\geq 3$: $$\begin{eqnarray*}\log(n!) = \sum_{k=1}^{n}\log(k) &=& n\log(n) - \sum_{k=1}^{n-1}k\log\left(1+\frac{1}{k}\right)\\&\geq& n\log n-\sum_{k=1}^{n}k\cdot\frac{1}{k}\\&=&n\left(\log n-1\right)\tag{1}\end{eqnarray*} $$ since $\log(1+x)\leq x$ for any $x>0$. The $p$-test hence ensures that $\sum_{n\geq 2}\frac{1}{\log^2(n!)}$ is converging, but the sum has to avoid the $n=1$ term since $\log(1!)=0$.
Jack D'Aurizio
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