Hardy- Littlewood Circle Method I'm currently trying to get to grips with the Hardy Littlewood circle method so I'm working through Vaughan's book.
In the past I've been very bad for leaving a point behind if I don't follow it so I'm not trying to get rid of that habit by asking more questions.
For example on page 3 of his book (see page here) Vaughan tells us that for 
$F(z)= \sum_{m=1}^{\infty} z^{a_m}$
if $a_m=m^2,\  \rho=1- \frac{1}{n}$, with n large and $e(\alpha)= e^{2 \pi i \alpha} $ then the function F has 'peaks' when $z= \rho e(\alpha)$ is 'close' to the point $e(a/q)$ with $q$ 'not too large'.
So (although I accept that this sounds stupid) I'm not sure about what he means by peaks, does he simply mean F has a peak or? and also why do the values of $z$ close the point $e(a/q)$ cause this?
He then goes on to tell us that F has an asymptotic expansion in the neighbourhood of such points, roughly speaking valid when $|\alpha - a/q| \leq 1/(q \sqrt{n})$ and $ q \leq \sqrt{n}$.  
Here I'm not sure what he means by asymptotic expansion and why that neighbourhood is valid.
Any help with any of this would be greatly appreciated. 
 A: $F$ is a complex-valued function, so you can think of "peaks" as local maxima of $|F(\rho e(\alpha))|$, where $0 < \rho < 1$ is fixed.  You can imagine that these peaks happen around certain rational values $\alpha = a/q$ ($q$ not too large) because the oscillations of $e(\alpha)$ are "in phase" with each other at rational points of the $[0,1)$ interval.  As $\rho$ gets closer to 1, the local peaks become more frequent and pronounced.
Vaughan mentions the asympotic expansion:
$$F\left(\rho e\left(\frac{a}{q}+\beta\right)\right) \sim \frac{C}{q} S(q,a)(1 - \rho(\beta))^{-1/2}$$
where $n$ is large, $\rho = 1 - 1/n$ and
$$S(q,a) = \sum_{m=1}^q e(am^2/q).$$
Vaughan says the asymptotic expansion works for denominator $q \leq \sqrt{n}$ and $\beta$ small, roughly $\beta \leq 1/(q\sqrt{n})$.  You can interpret this to mean that $F(\rho e(a/q+\beta))$ is approximately equal to $\frac{C}{q} S(q,a)(1 - \rho(\beta))^{-1/2}$ for $\rho$ close to 1 and $q, \beta$ in the ranges given, where $C$ here is actually $\sqrt{\pi}/2$.
In fact the asymptotic estimate does not seem to quite work on the full range of $q$ and $\beta$ that Vaughan gives, and I think we need to be somewhat more restrictive.  I give more explicit estimates below.  Perhaps somewhat else can give better estimates that allow the asymptotic to work for a larger range.
First, we'll need an application of partial summation.  Suppose that $f:\mathbb{R}_{\geq 0} \rightarrow \mathbb{C}$ is continuous and $f(x) \rightarrow 0$ as $x \rightarrow \infty$.  Then
$$\sum_{n=1}^\infty f(n) = \int_0^\infty f(x) \, dx + \int_0^\infty \{x\} f'(x) \,dx$$
where $\{x\}$ is the fractional part of $x$.  More generally, if we have a congruence condition for our summation, we have the estimate
$$\sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty f(n) = \frac{1}{q} \int_0^\infty f(x) \, dx - \int_0^\infty f(x) \,d\left\{\frac{x-m}{q}\right\}$$
$$= \frac{1}{q} \int_0^\infty f(x) \, dx - f(0)\left(1 - \frac{m}{q}\right) + \int_0^\infty \left\{\frac{x-m}{q}\right\} f'(x) \,dx.$$
Therefore
$$\left| \sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty f(n) -  \frac{1}{q} \int_0^\infty f(x) \, dx \right| \leq |f(0)| + \int_0^\infty |f'(x)| \, dx.$$
Let's apply this to the function
$$f(x) = \left(\rho e(\beta) \right)^{x^2}.$$
We have
$$\left| \sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty \left(\rho e(\beta) \right)^{n^2} - \frac{1}{q} \int_0^\infty \left(\rho e(\beta) \right)^{x^2} \, dx\right| \leq 1 + \int_0^\infty \left|\left(\left(\rho e(\beta) \right)^{x^2}\right)'\right| \, dx.$$
We evaluate the derivatives and integrals to get
$$\left| \sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty \left(\rho e(\beta) \right)^{n^2} - \frac{\sqrt{\pi}}{2q\sqrt{-\log{\rho e(\beta)}}}\right| \leq 1 + \int_0^\infty \left| 2x\left(\left(\rho e(\beta) \right)^{x^2}\right)(\log \rho + 2 \pi i \beta)\right| \, dx$$
$$\leq 1 + (-\log \rho + 2 \pi \beta)\int_0^\infty 2x\rho^{x^2} \, dx = 2 + \frac{2 \pi \beta}{(-\log{\rho})}.$$
Now we can estimate the full sum.  We split into residue classes to get
$$\sum_{n=1}^\infty \left(\rho e(a/q + \beta) \right)^{n^2} = \sum_{m (\bmod q)} e(am^2/q) \sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty \left(\rho e(\beta) \right)^{n^2}.$$
Therefore we have an estimate for the full sum
$$\left|F\left(\rho e\left(\frac{a}{q}+\beta\right)\right) - \frac{\sqrt{\pi}/2}{q}  S(q,a) \frac{1}{\sqrt{-\log \rho e(\beta)}}\right| \leq \left(2 + \frac{2 \pi \beta }{(-\log \rho)}\right) q.$$
The asymptotic estimate will be with fixed modulus $q$ and sending $\rho \rightarrow 1^-$.  Specifically, we fix $q$ and a small $\epsilon > 0$, and put $\rho = 1 - 1/n$.  We send $n \rightarrow \infty$, so that $\rho \rightarrow 1^-$.  At the same time, for each $n$ we need to choose $\beta$ such that $\beta = O(n^{-2/3-\epsilon})$, so $\beta$ goes to zero faster than Vaughan indicated.
Note that for small $\rho$ and $\beta$, we have
$$-\log \rho e(\beta) \sim 1- \rho e(\beta)  = O\left(\frac{1}{n^{2/3+\epsilon}}\right).$$
We thereby obtain an estimate
$$F\left(\rho e\left(\frac{a}{q}+\beta\right)\right) = \frac{\sqrt{\pi}/2}{q}  S(q,a) \frac{1}{\sqrt{1- \rho e(\beta)}} + O(n^{1/3 - \epsilon}).$$
Since $(1- \rho e(\beta))^{-1/2} \gg n^{1/3 + \epsilon/2} \gg n^{1/3 - \epsilon}$, we get the asymptotic
$$F\left(\rho e\left(\frac{a}{q}+\beta\right)\right) \sim \frac{\sqrt{\pi}/2}{q}  S(q,a) \frac{1}{\sqrt{1- \rho e(\beta)}}$$
as $n \rightarrow \infty$.
