Upper bound $f(t,d)$ such that $\sum_{j=1}^{d} cos(2 \pi j \; t) \leq f(t,d)$? I have a sum of a series of trig function as follows:
$\sum_{j=1}^{d} cos(2 \pi j \; t)$ where t is just a constant.
I am looking for the upper bound $f(t,d)$ such that $\sum_{j=1}^{d} cos(2 \pi j \; t) \leq f(t,d)$. Any idea or link or some papers that I can find hint about this problem will be much appreciated! Thanks ;) 
Here,  we can assume $t$ is a small number and $t \neq 0$.
Thanks a lot!
 A: All the cosines are bounded above by 1, so $d$ is an upper bound. It is tight because at $t=0$ the sum has value $d$.

In fact, we can do this sum exactly: multiplying by $\sin{\pi t}$, and using the prosthaphaeresis formula
$$ 2\cos{A}\sin{B} = \sin{(A+B)}-\sin{(A-B)} $$
we have
$$ \sum_{n=1}^d \cos{2\pi n t}\sin{\pi t} = \frac{1}{2}\sum_{n=1}^d \left(\sin{(2n+1)\pi t}-\sin{(2n-1)\pi t}\right),  $$
which we can see telescopes since $2(n+1)-1=2n+1$, and hence we only have the end terms, which are $\sin{(2d+1)\pi t}$ and $-\sin{\pi t}$, giving the result
$$ \sum_{n=1}^d \cos{2\pi n t} = \frac{\sin{(2d+1)\pi t}}{2\sin{\pi t}}-\frac{1}{2}.  $$
Now, to get an upper bound near $t=0$, we can expand this in a power series: for $0<x<\sqrt{6}$, we have
$$ x-x^3/6<\sin{x}<x, $$
from which we have
$$ \frac{1}{\pi t} \leqslant \frac{1}{\sin{\pi t}} \leqslant \frac{1}{\pi t-(\pi t)^3/6} = \frac{1}{\pi t} \frac{1}{1-\pi^2 t^2/6}, $$
for $\lvert t \rvert <\sqrt{6}/\pi$,  and
$$ \sin{(2d+1)\pi t} \leqslant (2d+1)\pi t, $$
so the whole lot is less than
$$ \frac{(2d+1)\pi t}{2\pi t}-\frac{1}{2} = d $$
For a less trivial estimate, probably your best bet is keeping the top sine intact, and taking
$$ \frac{\sin{(2d+1)\pi t}}{2\pi t(1-\pi^2 t^2/6)}-\frac{1}{2} $$
as the upper limit.
On the other hand, there are the bounds $x>\sin{x}>\frac{2}{\pi} x$ for $0<x<\pi/2$, which may be of use to you.
For a lower bound, you can expand the Taylor series around $t=0$ up to the negative quadratic term, which is
$$\begin{align*}
-&\frac{1}{2}+\frac{(2d+1)\pi t(1-(2d+1)^2\pi^2 t^2/6)+O(t^5)}{2\pi t(1-\pi^2 t^2/6)+O(t^5)} \\
&= -\frac{1}{2}+(d+1/2)\left( 1-\frac{(2d+1)^2\pi^2}{6} t^2+O(t^5) \right) \left( 1+\frac{\pi^2}{6} t+O(t^5) \right) \\
&= d + \frac{\pi^2}{12}(2d+1) \left( 1-(2d+1)^2 \right) t^2 + O(t^4) \\
&= d -\frac{\pi^2}{3}d(d+1)(2d+1) t^2 + O(t^4)
\end{align*}$$
(Now, notice the quadratic term looks suspiciously like the sum of the squares? We know perfectly well why that is...)
A: (Nothing really original here,
though I know my use of
$\mathbb{Re}$
for "real part" is wrong.)
$\begin{array}\\
s(d, t)
&=\sum_{j=1}^{d} cos(2 \pi j \; t)\\
&=\sum_{j=1}^{d} \mathbb{Re} e^{2 \pi ij t}\\
&=\mathbb{Re}\sum_{j=1}^{d}  e^{2 \pi ij t}\\
&=\mathbb{Re}e^{2\pi it}\sum_{j=0}^{d-1}  e^{2 \pi ij t}\\
&=\mathbb{Re}e^{2\pi it}\dfrac{e^{2 \pi id t}-1}{e^{2 \pi i t}-1}\\
&=\mathbb{Re}e^{2\pi it}\dfrac{(e^{2 \pi id t}-1)(e^{-2 \pi i t}-1)}{(e^{2 \pi i t}-1)(e^{-2 \pi i t}-1)}\\
&=\mathbb{Re}e^{2\pi it}\dfrac{e^{2 \pi i(d-1) t}-e^{2 \pi id t}-e^{2 \pi i t}+1}{1-(e^{-2 \pi i t}+e^{2 \pi i t})+1)}\\
&=\mathbb{Re}\dfrac{e^{2 \pi id t}-e^{2 \pi i(d+1) t}-e^{4 \pi i t}+e^{2\pi it}}{2-2\cos(2 \pi t)}\\
&=\dfrac{\cos(2 \pi d t)-\cos(2 \pi (d+1) t)-\cos(4 \pi  t)+\cos(2\pi t)}{2(1-\cos(2 \pi t))}\\
&=\dfrac{\cos(2 \pi d t)-\cos(2 \pi (d+1) t)-\cos(4 \pi  t)+\cos(2\pi t)}{2(2\sin^2( \pi t))}\\
&=\dfrac{\cos(2 \pi d t)-\cos(2 \pi (d+1) t)-\cos(4 \pi  t)+\cos(2\pi t)}{4\sin^2( \pi t)}\\
\end{array}
$
Since
$2t
<\sin(\pi t)< \pi t
$
(because
$\sin(x) > \frac{2x}{\pi}$
for $0 < x < \frac{\pi}{2}$),
and all the $\cos$ terms are
at most $1$,
$|s(d, t)|
\le \frac1{\sin^2(\pi t)}
<\frac1{4t^2}
$.
You can get better bounds
that actually involve $d$
by looking at the differences
$\cos(2 \pi d t)-\cos(2 \pi (d+1) t)$
and
$\cos(4 \pi  t)-\cos(2\pi t)$,
but I'll leave that to you.
