# Tagged Questions

119 views

### Showing $n!<e(\frac{n}{2})^n$

I'd like to prove that $n!<e(\frac{n}{2})^n$. What I have so far: $\sqrt[n]{n!} = \sqrt[n]{1\cdot 2 \cdot \ldots \cdot n} \leq \frac{1+\ldots +n}{n}=\frac{(n+1)n}{2n}=\frac{(n+1)}{2}$. Thus ...
136 views

### Study of the convergence of a sequence with repeated radicals

Let the sequence $$a_n = \sqrt {1!\sqrt {2!\cdots\sqrt {n!} } }$$ Does this sequence converge? I can tell intuitively that $a_n$ is monotonically increasing. Therefore, there are two ...
50 views

### Which series of numbers effectively translates the factorial to the exponential function?

We have the relation of the Bernoulli numbers $$B_{2n} = (-1)^{n+1}\frac {2(2n)!} {(2\pi)^{2n}} \left(1+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\cdots\;\right).$$ For $n>1$, the right hand sum ...
64 views

### Stirling approximation / Gamma function

Is it possible to obtain the Stirling approximation for the factorial by using the gamma function ? $$\Gamma(z) = \lim_{n \to +\infty} \frac{n! n^{z}}{z(z+1) \dots (z+n)}$$ Any hint ?
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### Limit of the sequence $\frac {a^n} {n!}$

I need to prove that $$\lim_{n \rightarrow \infty} \frac {a^n} {n!}=0$$ I have no condition over $a$, just that is a real number. I thought of using L'HÃ´pital, but it's way too complicated for ...
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### limit of sequence with factorial

How do you show that: $\lim\limits_{n\to \infty} \frac{\left(\frac{n}{2}\right)^{\frac{n}{2}}}{n!}=0$ using the squeeze theorem (I'd like to avoid using Stirling's formula, too). I tried rearranging ...
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### Summation of a curious series-repeated division by primes

I am interested in knowing if there is some closed form/formula for the following series: ...
150 views

### Infinite Sum of factorial denominator and exponential numerator

I've been trying to find the sum of the following infinite series: $$\sum\limits_{n=1}^\infty \frac{x^n}{n!2^n}$$ I've rewritten it as $$\sum\limits_{n=1}^\infty \frac{y^n}{n!}, y=\frac{x}{2}$$ ...
### Why doesn't $0! = 1$ in the context of this general term?
Is my instructor wrong to say that $\left\{0,\frac{1!}{4},\frac{2!}{9},\frac{3!}{16},\dots\right\} = \left\{\frac{(n-1)!}{n^2}\right\}$? My understanding is that at $n=1$, $\frac{(n-1)!}{n^2}$ should ...