Use the Geometric Series Theory, we know that $\sum\frac{1}{4^n}$ convergence, because $\sum\frac{1}{4^n} = \sum\frac{1}{4} \cdot {\frac{1}{4}^{n-1} }$ and $\frac{1}{4} < 1$.
Last week, we’ve learned Comparison Test Theory. So if we can find a series that every element in it is bigger than this one and also converges, then this example must converge, too.
So if n is big enough, then the common ratio will be close to 1/4.
Then we can pick a constant $\epsilon = 0.1$ that make $\vert \frac{a_n}{a_{n-1} } \vert < \frac{1}{4} + 0.1$, after some calculation we got that $n \ge 15$.
So, $a_{n+1} < (\frac{1}{4} + 0.1) a_n,\\ \text{where}\\ n \ge 15$,
Then $a_{15+k} < (\frac{1}{4} + 0.1)^k a_{15}$.
We know that $\sum (\frac{1}{4} + 0.1)^k a_{15}$ convergence, base on the Geometric Series Theory.
We know that $e = \displaystyle\lim_{n \to \infty} (1+\frac{1}{n})^n$. With this, we can conclude $\sum \frac{n!}{(n/2)^n}$ converges, $\sum \frac{n!}{(n/3)^n}$ diverges.
But what about $\sum \frac{n!}{(n/e)^n}$. Let’s compare $n!$ with $n^n$
$\log(n!) = \sum_{k=1}^n \log(k) \approx \int_1^n \log x dx = x \log x - x ]_1^n$
$\log(n!) = n \log n - n + 1$
$\log(n!) \approx n \log n - n = \log(\frac{n^n}{e^n})$
$n! \approx (\frac{n}{e})^n$
Better approximation is $n! \approx (\frac{n}{e})^n \sqrt{2 \pi n}$, which name is Stirling’s approximation.
Suppose f is a continuous, positive, decreasing function on $[1, \infty)$, and let $a_n = f(n)$. Then the series $\sum_{n=1}^{\infty} a_n$ is convergent if and only if the improper integral $\int_{1}^{\infty} f(x) dx$ is convergent. In other words:
If $\int_{1}^{\infty} f(x) dx$ is convergent, then $\sum_{n=1}^{\infty} a_n$ is convergent.
If $\int_{1}^{\infty} f(x) dx$ is divergent, then $\sum_{n=1}^{\infty} a_n$ is divergent.
For the general series $\sum a_n$, look at the figures above. The area of the first shaded rectangle is the value of f at the right endpoint of [1, 2], that is, $f(2) = a_2$. So, comparing the areas of the shaded rectangles with the area under $y = f(x)$ from 1 to n, we see that
Notice that this inequality depends on the fact that f is decreasing.
If $\int_1^n f(x) dx$ is convergent, then $$\sum_{i=2}^n a_i \le \int_1^n f(x) dx \le \int_1^{\infty} f(x) dx$$ since $f(x) \ge 0$. Therefore $$s_n = a_1 + \sum_{i=2}^n a_i \le a_1 + \int_1^{\infty} f(x) dx$$
So the sequence ${s_n}$ is bounded above.
And ${s_n}$ is also an increasing sequence.
This means that $\sum a_n$ is convergence.
If $\int_1^n f(x) dx$ is divergent, then $\int_1^n f(x) dx \to \infty \\ \text{as}\\ n \to \infty$ because $f(x) \ge 0$. But $\displaystyle\int_1^n f(x) dx \le \sum_{i=1}^{n-1}a_i = s_{n-1}$ and so $s_{n-1} \to \infty$. This implies that $s_n \to \infty$ and $\sum a_n$ diverges.
With Cauchy Condensation, we know that, $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if and only if $\sum_{n=0}^{\infty} 2^n \frac{1}{(2^n)^p}$ converges.
