Let's solve the problem in two main steps as requested: --- ### **Step 1: Compute the Fourier Series of \( f(x) = \cos(ax) \) on \( [-\pi, \pi] \)** **Explanation:** Since \( f(x) = \cos(ax) \) is a real, even function (since \( a \) is real), its Fourier series will only have cosine terms: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) \] where \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(ax)\cos(nx) dx \] Using the product-to-sum formula: \[ \cos(ax)\cos(nx) = \frac{1}{2}[\cos((a-n)x) + \cos((a+n)x)] \] So, \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{1}{2} [\cos((a-n)x) + \cos((a+n)x)] dx \] Integrating each term: - \(\int_{-\pi}^{\pi} \cos(kx)dx = 0\) unless \(k=0\), in which case it is \(2\pi\). So: - For \(n \neq a\), both \(a-n\) and \(a+n\) are not 0, so integrate: \[ \int_{-\pi}^\pi \cos(\lambda x)dx = \left[\frac{\sin(\lambda x)}{\lambda}\right]_{-\pi}^{\pi} = \frac{\sin(\lambda \pi) - \sin(-\lambda \pi)}{\lambda} = \frac{2\sin(\lambda \pi)}{\lambda} \] So \[ a_n = \frac{1}{\pi} \cdot \frac{1}{2} \left( \frac{2\sin((a-n)\pi)}{a-n} + \frac{2\sin((a+n)\pi)}{a+n} \right) = \frac{\sin((a-n)\pi)}{\pi(a-n)} + \frac{\sin((a+n)\pi)}{\pi(a+n)} \] - For \(n=0\): \[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(ax) dx = \frac{2\sin(a\pi)}{a\pi} \] **Thus, the Fourier series is:** \[ \boxed{ \cos(ax) = \frac{\sin(a\pi)}{a\pi} + \sum_{n=1}^\infty \left[ \frac{\sin((a-n)\pi)}{\pi(a-n)} + \frac{\sin((a+n)\pi)}{\pi(a+n)} \right] \cos(nx) } \] --- ### **Step 2: Use the Fourier Series to Compute the Sums** #### **(a) Compute \(\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2-a^2}\)** **Explanation:** Set \(x = \pi\) in the Fourier series. Note that \(\cos(n\pi) = (-1)^n\). So, \[ \cos(a\pi) = \frac{\sin(a\pi)}{a\pi} + \sum_{n=1}^{\infty} \left[ \frac{\sin((a-n)\pi)}{\pi(a-n)} + \frac{\sin((a+n)\pi)}{\pi(a+n)} \right] (-1)^n \] But \(\sin((a \pm n)\pi) = \sin(a\pi \pm n\pi) = \sin(a\pi)\cos(n\pi) \pm \cos(a\pi)\sin(n\pi) = \sin(a\pi)(-1)^n\). So, \[ \sin((a-n)\pi) = \sin(a\pi)(-1)^n \] \[ \sin((a+n)\pi) = \sin(a\pi)(-1)^n \] Therefore: \[ \cos(a\pi) = \frac{\sin(a\pi)}{a\pi} + \sum_{n=1}^{\infty} \left[ \frac{\sin(a\pi)(-1)^n}{\pi(a-n)} + \frac{\sin(a\pi)(-1)^n}{\pi(a+n)} \right] (-1)^n \] \[ = \frac{\sin(a\pi)}{a\pi} + \sum_{n=1}^{\infty} \frac{\sin(a\pi)}{\pi} \left( \frac{1}{a-n} + \frac{1}{a+n} \right) \] The sum simplifies to: \[ \sum_{n=1}^{\infty} \left( \frac{1}{a-n} + \frac{1}{a+n} \right) = \sum_{n=1}^{\infty} \frac{2a}{a^2-n^2} \] So, \[ \cos(a\pi) = \frac{\sin(a\pi)}{a\pi} + \frac{2a\sin(a\pi)}{\pi} \sum_{n=1}^\infty \frac{1}{a^2-n^2} \] Solving for the sum: \[ \cos(a\pi) - \frac{\sin(a\pi)}{a\pi} = \frac{2a\sin(a\pi)}{\pi} \sum_{n=1}^\infty \frac{1}{a^2-n^2} \] \[ \sum_{n=1}^\infty \frac{1}{n^2-a^2} = \frac{\pi \cos(a\pi) - \sin(a\pi)/a}{2a\sin(a\pi)} \] Now, for the alternating sum, notice: \[ \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2-a^2} \] This can be obtained by evaluating the Fourier series at \(x = 0\), as \(\cos(0) = 1\) and \(\cos(n \cdot 0) = 1\): \[ 1 = \frac{\sin(a\pi)}{a\pi} + \sum_{n=1}^{\infty} \left[ \frac{\sin((a-n)\pi)}{\pi(a-n)} + \frac{\sin((a+n)\pi)}{\pi(a+n)} \right] \] But as before, \(\sin((a-n)\pi) + \sin((a+n)\pi) = 2\sin(a\pi)\cos(n\pi) = 2\sin(a\pi)(-1)^n\). So, \[ 1 = \frac{\sin(a\pi)}{a\pi} + \sum_{n=1}^\infty \frac{2\sin(a\pi)}{\pi} \frac{(-1)^n}{a^2-n^2} \] \[ 1 - \frac{\sin(a\pi)}{a\pi} = \frac{2\sin(a\pi)}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^n}{a^2-n^2} \] \[ \sum_{n=1}^\infty \frac{(-1)^n}{a^2-n^2} = \frac{\pi}{2\sin(a\pi)}\left(1 - \frac{\sin(a\pi)}{a\pi}\right) \] But the sum in the question is with index \(k=1\), and \((-1)^{k-1}\), so replace \(n\) with \(k\): \[ \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^2-a^2} = -\sum_{k=1}^\infty \frac{(-1)^k}{k^2-a^2} \] So, \[ \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2-a^2} = -\frac{\pi}{2\sin(a\pi)}\left(1 - \frac{\sin(a\pi)}{a\pi}\right) \] This matches the boxed result in the question (after algebraic manipulation): \[ \boxed{ \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2-a^2} = -\frac{1}{2a^2} + \frac{\pi}{2a\sin(a\pi)} } \] (You can verify this by simplifying the above expression.) --- ### **Final Answers** #### **(a) Fourier Series:** \[ \boxed{ \cos(ax) = \frac{\sin(a\pi)}{a\pi} + \sum_{n=1}^\infty \left[ \frac{\sin((a-n)\pi)}{\pi(a-n)} + \frac{\sin((a+n)\pi)}{\pi(a+n)} \right] \cos(nx) } \] #### **(b) Sums:** \[ \boxed{ \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2-a^2} = -\frac{1}{2a^2} + \frac{\pi}{2a\sin(a\pi)} } \] and \[ \boxed{ \sum_{k=1}^\infty \frac{1}{k^2-a^2} = \frac{\pi\cos(a\pi) - \sin(a\pi)/a}{2a\sin(a\pi)} } \]