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I have found that the limit of x to infinity of x!/x^x is zero by doing this:

x^x = xxxx...(x amount of times) x! = x(x-1)(x-2)(x-3)...(x amount of times)

So obviously, x^x grows much faster, because x is greater than x-1 and x-2 and x-3 and all those other terms. Therefore the limit must be 0.

But how to actually prove this with equations. And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:

https://www.wolframalpha.com/input/?i=sum+x!%2Fx%5Ex,+j%3D0..infinity

Math Bob
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    The limit of a sequence and the sum of a series are different things ... – Martin R Jan 02 '19 at 19:34
  • Are you calculating the limit of that sequence or the series of the sequence?? –  Jan 02 '19 at 19:34
  • Regarding the WA link, note that you write x!/x^x, but sum over j... – Wojowu Jan 02 '19 at 19:35
  • The reason that wolfram alpha says your sum diverges is as Wojowu says you used the variable $x$ instead of $j$ in that sum. On top of this you need to start the index at $j=1$ on wolfram because of its convention with $0^0$. you can see this here: https://www.wolframalpha.com/input/?i=sum+j!%2Fj%5Ej,+j%3D1..infinity Contrary to the answer youve accepted the sum actually does converge, and you can show this pretty easily with a comparison with $\sum \frac{2}{n^2}$ – Ben Martin Jan 03 '19 at 00:19

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Split it in terms of the form $\frac{n}{x}$, with $n$ varying from $1$ to $x$. Clearly this is bounded above by the first term, which is $\frac{1}{x}$, which is null, so the sequence tends to $0$.

And why does Wolfram Alpha disagree with me when I do the infinite summation of x!/x^x and it says by the limit test that the summation diverges:

Because, while the sequence $$\left(\dfrac{x!}{x^x}\right)$$ converges, the series $${\sum}\dfrac{x!}{x^x}$$ does not.

user3482749
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  • But according to Khan Academy here: https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/modal/v/integral-test-intuition, the improper integral of a function is an upper bound for its infinite summation, proving a very good argument for its convergence. On Wolfram, when I integrate x!/x^x from 0 to infinity I get something around 2.51791766302214. Shouldn't that be an upper bound for the series' summation, proving its convergence? – Math Bob Jan 02 '19 at 20:48