I'm reading Zygmund's Trigonometric Series (precisely Lemma 6.6 of chapter 12), and I'm struggling to understand the following detail. $(a_n)$ here is a sequence of positive numbers and decreasing. Also, $0<x<\pi$.
By using a corollary of summation by parts (he explicitly refers to that result, 2.3 of chapter 1) , he states that $$\left|\sum_{\nu=n+1}^{\infty}a_\nu cos(\nu x)\right|\leq \frac{\pi a_n}{x} $$
I really can't see how he gets this inequality. Any ideas?
So, applying the summation by parts on $\left|\sum_{v=n+1}^\infty a_v \cos(vx)\right|$ it implies: $$ \left|\sum_{v=n+1}^\infty a_v \cos(vx) \right| \leq a_{n+1}\cdot h $$ Where $h$ is a bound for the partial sums of $\sum_{v=n+1}^N \cos(vx)$. $$ \sum_{v=n+1}^N \cos(vx) = \sum_{v=0}^M \cos(vx + (n+1)x) = \frac{\sin\frac{(M+1) x}{2} \cdot \cos\left((n+1)x + \frac{M x}{2}\right)}{\sin\frac{x}{2}} \leq \frac{1}{\sin \frac{x}{2}} $$ For now we get $$ \left|\sum_{v=n+1}^\infty a_v \cos(vx) \right| \leq \frac{a_{n+1}}{\sin \frac{x}{2}} \leq \frac{a_{n}}{\sin \frac{x}{2}} $$
In the case were $\pi/2>x>0$, we have bounds $x>\sin(x)>x/(\pi/2)$ (that can be easily seen from the graph of $\sin(x)$ or, more formally, through Jensen's inequality), thus: $$ \sin\left(\frac{x}{2}\right)>\frac{\left(x/2\right)}{\pi/2} = \frac{x}{\pi} $$ So, $$ \left|\sum_{v=n+1}^\infty a_v \cos(vx) \right| \leq \frac{a_{n}}{\sin \frac{x}{2}} \leq \frac{a_{n}}{x/\pi}=\frac{\pi~a_{n}}{x} $$