$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$

Question

I have to study the character of this series $$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$$ and I tried with the ratio test:

$ \frac{a_{n+1}}{a_n}= \frac{((n+1)!)^{-\frac{1}{n+1}}}{(n!)^{-\frac{1}{n}}}= \frac{(n!)^{\frac{1}{n}}}{((n+1)!)^{\frac{1}{n+1}}}= \frac{(n!)^{\frac{1}{n}}}{((n+1) \star n!)^{\frac{1}{n+1}}}= \frac{(n!)^{\frac{1}{n}-\frac{1}{n+1}}}{((n+1) )^{\frac{1}{n+1}}} \sim (n!)^{\frac{1}{n}-\frac{1}{n+1}}= (n!)^{\frac{1}{n(n+1)}}>(n(n+1))^{\frac{1}{n(n+1)}} \sim 1$

So the limit is larger than 1 and the given series diverges.

Is it right?

Just because it's larger than something that goes to $1$ doesn't mean it goes to something larger than $1$. — saulspatz, Mar 08 '20 at 16:08
The limit you're evaluating is, in fact, $1$, so the ratio test doesn't work here. — bjorn93, Mar 08 '20 at 16:24
Such formulas can be simplified by taking logarithms. Perhaps a judicious use of bounds on factorials allows answering this — vonbrand, Mar 08 '20 at 18:17

score 2 · Answer 1 · answered Mar 08 '20 at 16:50

2

$\frac{1}{n}=\frac{1}{n^{n*\frac{1}{n}}}<\frac{1}{n!^{\frac{1}{n}}}$

Applying the comparison test and knowing that the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, also our series diverges

answered Mar 08 '20 at 16:50

Anne

2,931

$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$

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