Let a_n = int_{-pi}^{pi} |x-1| cos(nx) , dx f | vedclass

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(D) By the Riemann-Lebesgue Lemma,if $f(x)$ is an integrable function on $[a, b]$,then $\lim_{n \rightarrow \infty} \int_{a}^{b} f(x) \cos(nx) \, dx =

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$\lim_{n \rightarrow \infty} a_n = 0$

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(D) By the Riemann-Lebesgue Lemma,if $f(x)$ is an integrable function on $[a, b]$,then $\lim_{n \rightarrow \infty} \int_{a}^{b} f(x) \cos(nx) \, dx = 0$. Here,$f(x) = |x-1|$ is a continuous and therefore integrable function on the interval $[-\pi, \pi]$. Since $f(x) = |x-1|$ is integrable,the integral $a_n = \int_{-\pi}^{\pi} |x-1| \cos(nx) \, dx$ represents the Fourier cosine coefficient (up to a constant factor) of the function $f(x)$. According to the Riemann-Lebesgue Lemma,as $n \rightarrow \infty$,the integral of a function multiplied by $\cos(nx)$ approaches $0$. Therefore,$\lim_{n \rightarrow \infty} a_n = 0$.

Let $a_n = \int_{-\pi}^{\pi} |x-1| \cos(nx) \, dx$ for all natural numbers $n$. Then,the sequence $(a_n)_{n \geq 1}$ satisfies:

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