Question

Using the Fourier series, the value of
\[ \sum_{n=0}^{\infty} \frac{1}{(2n-1)^{2}} \]
is:

• A. \( \dfrac{1}{2} \)
• B. \( \dfrac{\pi^{2}}{8} \)
• C. \( \dfrac{\pi}{8} \)
• D. \( \dfrac{\pi^{2}}{2} \)

Hint:

The odd reciprocal squares sum to \( \pi^2/8 \). Get it from the Fourier series of \( |x| \) at \( x=0 \), or subtract the even part \( \pi^2/24 \) from Basel's \( \pi^2/6 \).

Answer 1

The Correct Option is B

Step 1: The sum is the reciprocal-squares of the odd integers, \( \dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+\cdots \). We evaluate it from a Fourier series.

Step 2: Expand \( f(x)=x \) on \( (-\pi,\pi) \) is odd; instead take the standard series for \( f(x)=|x| \) on \( (-\pi,\pi) \):
\[ |x| = \frac{\pi}{2} - \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\cos\big((2k+1)x\big)}{(2k+1)^{2}}. \]

Step 3: Put \( x=0 \). Then \( |x|=0 \) and \( \cos 0 = 1 \), giving
\[ 0 = \frac{\pi}{2} - \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2}}. \]

Step 4: Solve for the sum:
\[ \frac{4}{\pi}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2}} = \frac{\pi}{2} \;\Rightarrow\; \sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2}} = \frac{\pi^{2}}{8}. \]

Step 5: The odd terms \( (2n-1) \) for \( n=1,2,3,\dots \) are exactly \( 1,3,5,\dots \), the same set, so
\[ \sum \frac{1}{(2n-1)^{2}} = \frac{\pi^{2}}{8}. \]
\[ \boxed{\, \dfrac{\pi^{2}}{8} \,} \]