$B_N(x)=\sum_{n=1}^N\cos nx$

$$\sin(A+B)-\sin(A-B)=\sin A\cos B-\cos A\sin B-\sin A\cos B+\cos A\sin B=2\sin B\cos A$$

$$\sin(x/2)\sum_{n=1}^N\cos(nx)=\frac{1}{2}\sum_{n=1}^N\left(\sin(nx+x/2)-\sin(nx-x/2)\right)$$

כמעט הכל מצטמצם ונשאר:

$$2\sin(x/2)\sum_{n=1}^N\cos(nx)=\sin(Nx+x/2)-\sin(x-x/2)$$

$$\sum_{n=1}^N\cos(nx)=\frac{\sin(Nx+x/2)-\sin(x/2)}{2\sin(x/2)}$$

Convergence to zero of a function whose improper integral converges.

It is well known that if a series $\sum_{n=1}^{\infty}a_n$ converges, then $a_n\to 0$ as $n\to\infty$. The reason is that since the sequence of partial sums $S_N=\sum_{n=1}^Na_n$  is a convergent sequence of real numbers, it must be Cauchy, so in particular $|S_n|=|S_{n+1}-S_n|$ can be made smaller than any given $\varepsilon>0$, provided $n$ is sufficiently large.
A similar argument does not hold for convergent integrals. Contrary to this situation, if $f$ is a function for which the improper Riemann integral $\int_a^{\infty}f(x)\,dx$ converges, one cannot deduce that $\lim_{x\to\infty}f(x)=0$. Although one still has a Cauchy condition, namely, that the integrals $\int_{\alpha}^{\beta}f(x)\,dx$ are small if $\alpha,\beta$ are sufficiently large, there is enough freedom to form spikes on the graph of $f$, which do not contribute much to the integral. Such "spikes" will necessarily cause the function not to be uniformly continuous. If, however, the function is uniformly continuous, then $\lim_{x\to\infty}f(x)=0$, for if this were not the case, then for some $\varepsilon>0$ we could find an infinite sequence $\{x_n\}_{n=1}^{\infty}$ such that $|x_n-x_{n+1}|>1$, say, for which $|f(x_n)|>\varepsilon$ for all $n$. Since $f$ is uniformly continuous, there exists $0<\delta < 1$, such that for all $n$, if $|x-x_n|<\delta$, then $|f(x_n)-f(x)|<\varepsilon/2$. In particular,
$$\left| \int_{x_n-\delta}^{x_n+\delta}f(x)\,dx\right|\geq \delta\varepsilon\quad\hbox{for all $n$},$$
contradicting the Cauchy criterion.