In [the LPS paper "Ramanujan graphs"](http://www.ma.huji.ac.il/%7Ealexlub/PAPERS/ramanujan%20graphs/ramanujanGraphs.pdf) the adjacency matrix of $X^{p,q}$, for simplicity say that $p,q\equiv1\mod{4}$ and $\left(\frac{p}{q}\right)=1$ (so, nonbipartite) and $n=\lvert X^{p,q}\rvert$, is considered with spectrum $p+1=\lambda\_0>\lambda\_1\ge\cdots\ge\lambda\_{n-1}>-(p+1)$. Then, the "angle" $\theta\_j$ is introduced as $\lambda\_j=2\sqrt{p}\cos\theta\_j$, where
$$\theta\in\begin{cases}[0,\pi] & \text{$\lambda$ is in the bulk of the spectrum} \\\ i\mathbb{R}\_+ & \lambda>2\sqrt{p} \\\ \pi+i\mathbb{R} & \lambda<-2\sqrt{p}.\end{cases}$$
I think it is implicitly used that "$\cos$" is really $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$ and $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$, in order to allow for $\lambda$ outside of the bulk.


I am looking at Case ii. in the proof of Theorem 4.1, on pgs. 13/14 of the linked PDF version. The crucial expression ends up being (for even $k$)
$$\sum\limits\_{0\le j<n}\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}.$$
It is claimed without explanation that this sum equals
$$\frac{2}{p^{\frac{k}{2}}}\frac{p^{k+1}-1}{p-1}+o(p^\frac{k}{2}).$$
My first question is: **Where does this asymptotic evaluation come from?**


Then, the paper reaches the conclusion that
$$\sum\limits\_{0<j<n}\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}\in O(p^{k\epsilon})$$
(having already resolved the trivial eigenvalue) and claims then that $\theta\_j\in\mathbb{R}$ as a consequence. My second question is: **Why does this consequence hold?** My best guess for this is that for $\lambda$ not in the bulk, the corresponding $\theta$ has $t:=e^{i\theta}=\frac{\lambda}{2\sqrt{p}}\pm\sqrt{\frac{\lambda^2}{4p}-1}$ (the sign agreeing with $\lambda$'s) satisfying $\lvert t\rvert>1$ and we can compute that
$$\frac{\sin((k+1)\theta)}{\sin\theta}=\frac{1}{t^k}(1+t^2+t^4+\cdots+t^{2k})$$
so that for very large (still even) $k$ and small $\epsilon$, this positive term would dominate. However I am not sure how correct this is.



Series analyzed in Lubotzky–Phillips–Sarnak "Ramanujan Graphs"
**1.** Regarding your first question, we have $e^{i\theta\_0}=p^{-1/2}$ as emphasized two lines below (4.11). Therefore,
$$\frac{\sin((k+1)\theta\_0)}{\sin\theta\_0}=
\frac{e^{-i(k+1)\theta\_0}-e^{i(k+1)\theta\_0}}{e^{-i\theta\_0}-e^{i\theta\_0}}
=\frac{p^{(k+1)/2}-p^{-(k+1)/2}}{p^{1/2}-p^{-1/2}},$$
so that
$$\frac{2p^{k/2}}{n}\cdot\frac{\sin((k+1)\theta\_0)}{\sin\theta\_0}=\frac{2(p^{k+1}-1)}{n(p-1)}.\tag{1}$$
In contrast, for $j\in\{1,\dots,n-1\}$ we have
$$|e^{i\theta\_j}+e^{-i\theta\_j}|=2|\cos\theta\_j|=|\lambda\_j/\sqrt{p}|<p^{1/2}+p^{-1/2},$$
which eventually yields that $p^{-1/2}<|e^{i\theta\_j}|\leq 1$ by the line below (4.11). So in this case
$$\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}=
\frac{e^{-i(k+1)\theta\_j}-e^{i(k+1)\theta\_j}}{e^{-i\theta\_j}-e^{i\theta\_j}}
=o(p^{k/2}),\qquad k\to\infty,$$
whence also
$$\frac{2p^{k/2}}{n}\cdot\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}=o(p^k),\qquad k\to\infty.\tag{2}$$
Combining $(1)$ and $(2)$, we get
$$\frac{2p^{k/2}}{n}\sum\_{j=0}^{n-1}\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}=\frac{2(p^{k+1}-1)}{n(p-1)}+o(p^k),\qquad k\to\infty.$$
In the light of (4.16), this implies the display above (4.20).


**2.** Regarding your second question, let us start from
$$\sum\_{j=1}^{n-1}\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}=O(p^{k\epsilon}).$$
We can rewrite this as
$$\sum\_{j=1}^{n-1}\frac{e^{-i(k+1)\theta\_j}-e^{i(k+1)\theta\_j}}{e^{-i\theta\_j}-e^{i\theta\_j}}=O(p^{k\epsilon}).\tag{3}$$
Note that here $|e^{i\theta\_j}|\leq 1\leq|e^{-i\theta\_j}|$ by the line below (4.11). Let us record the smallest of these positive numbers as
$$r:=\min\left(|e^{i\theta\_1}|,\dots,|e^{i\theta\_{n-1}}|\right).$$
By $(3)$, the power series
$$f(z):=\sum\_{k=0}^\infty z^k\sum\_{j=1}^{n-1}\frac{e^{-i(k+1)\theta\_j}-e^{i(k+1)\theta\_j}}{e^{-i\theta\_j}-e^{i\theta\_j}}$$
converges in the open disk $D(0,p^{-\epsilon})$ for any $\epsilon>0$, hence also in the union of these disks, which is $D(0,1)$. In particular, $f(z)$ is a holomorphic function in $D(0,1)$. On the other hand, in the smaller open disk $D(0,r)$ we can evaluate $f(z)$ as
\begin{align\*}
f(z)&=\sum\_{j=1}^{n-1}\sum\_{k=0}^\infty\frac{z^k e^{-i(k+1)\theta\_j}-z^k e^{i(k+1)\theta\_j}}{e^{-i\theta\_j}-e^{i\theta\_j}}\\
&=\sum\_{j=1}^{n-1}\frac{e^{-i\theta\_j}(1-ze^{-i\theta\_j})^{-1}-e^{i\theta\_j}(1-ze^{i\theta\_j})^{-1}}{e^{-i\theta\_j}-e^{i\theta\_j}}\\
&=\sum\_{j=1}^{n-1}\frac{1}{(z-e^{i\theta\_j})(z-e^{-i\theta\_j})}.\end{align\*}
So the right-hand side extends holomorphically from $D(0,r)$ to $D(0,1)$, which means that its poles $e^{\pm i\theta\_j}$ cannot lie in $D(0,1)$. So each $e^{i\theta\_j}$ lies on the unit circle (rather than in the open unit disk), whence $\theta\_j\in\mathbb{R}$.


