I am currently reading Feller's "Introduction to Probabilty theory"... Please, can you help me to understand a part of his normal approximation to binomial?
For simplicity he took the case of $n=2v$ trials and $v+k$ successes, with $P(Success)=1/2$. Hence,
$a_k=Binom(v+k, 2v, 1/2)$, then he reasoned, that from the definiton
$a_k=a_0 \frac{v(v-1)...(v-k+1)}{(v+1)(v+2)...(v+k)}$
Then the part that I can't understand: He divided the numerator and denominator by $v^k$ and claimed that the individual factors take on the form $1 + j/v$ with $j$ running from $-(k-1)$ to $k$. Why did he divided and how it was done? He did it to get:
$1+ \frac{j}{v}=e^{(j/v)+...} $, which I understand, but how he came to this point?
In the equation connecting $a_n$ with $a_0$, the numerator $v(v-1)\cdots(v-k+1)$ has $k$ factors. (I'm writing $v$ instead of $\nu$ to save typing.) Feller divides each factor by its own copy of $v$, getting (after division) $(v/v) ((v-1)/v) \cdots((v-k+1)/v) = 1 (1-1/v) \cdots (1-(k-1)/v)$. Simlarly, the denominator $(v+1)(v+2)\cdots(v+k)$ has $k$ factors, each of which he divides by $v$ to obtain $((v+1)/v)((v+2)/v)\cdots ((v+k)/v) = (1+1/v)(1+2/v)\cdots(1+k/v)$.
Maybe it's clearer with a slight change of notation:$$ v (v-1) \cdots(v-k+1) =\prod_{j=0}^{k-1} (v-j)=\prod_{j=0}^{k-1} v\left(1-\frac j v\right)= v^k \prod_{j=0}^{k-1} \left(1-\frac j v\right)$$ for the numerator, and so on.
Why does he do this? It's a kind of factorization: both numerator and denominator are products of quantities close to the big quantity $v$, so he factors out the big $v^k$ from both numerator and denominator, leaving a product of numbers close to $1$.
Continuing, to fill in gaps suggested by the comments: the ratio $a_k/a_0$ is given by the product $$ \frac{a_k}{a_0} = \frac{\prod_{j=0}^{k-1} \left(1-\frac j v\right)}{\prod_{j=1}^{k} \left(1+\frac j v\right)}$$ $$=\exp\left( \sum_{j=0}^{k-1} \log\left(1-\frac j v\right)-\sum_{j=1}^{k}\log \left(1+\frac j v\right)\right)\approx\exp\left( \sum_{j=0}^{k-1} -\frac j v-\sum_{j=1}^{k}\frac j v \right). $$ Here the basic approximation is $\log(1+t)=t +O(t^2)$. After summation over $t=\pm j/v$ for $j=1,2,\dots k$, the error from the $O(t^2)$ term is $\sum_{j=1}^k O((j/v)^2) = O(k^3/v^2)$, which I'm sure Feller accounts for. (I'm writing from memory: my Feller is at the office just now.) The main term in the exponent is $$ \left(\sum_{j=0}^{k-1} -\frac j v\right)-\left(\sum_{j=1}^{k}\frac j v \right)= -\frac 1 v \left(\sum_{j=0}^{k-1} j + \sum_{j=1}^{k} j \right) = -\frac 1 v \left( \frac {k(k-1} 2 + \frac {k(k+1)}2\right) = - \frac {k^2} v.$$ This is why Feller approximates $a_k\approx a_0 \exp(-k^2/v)$.