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: Start from the well-known Basel result $\displaystyle \sum_{m=1}^{\infty}\frac{1}{m^{2}}=\frac{\pi^{2}}{6}$, which itself follows from the Fourier expansion of $x^{2}$.

Step 2: Split this into even and odd denominators:
\[ \sum_{m=1}^{\infty}\frac{1}{m^{2}} = \underbrace{\sum_{\text{even }m}\frac{1}{m^{2}}}_{S_e} + \underbrace{\sum_{\text{odd }m}\frac{1}{m^{2}}}_{S_o}. \]

Step 3: The even part factors as
\[ S_e = \sum_{k=1}^{\infty}\frac{1}{(2k)^{2}} = \frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^{2}} = \frac{1}{4}\cdot\frac{\pi^{2}}{6} = \frac{\pi^{2}}{24}. \]

Step 4: Therefore the odd part is
\[ S_o = \frac{\pi^{2}}{6} - \frac{\pi^{2}}{24} = \frac{4\pi^{2}-\pi^{2}}{24} = \frac{3\pi^{2}}{24} = \frac{\pi^{2}}{8}. \]

Step 5: Since $S_o$ is precisely $\sum 1/(2n-1)^{2}$, the required value is $\dfrac{\pi^{2}}{8}$, matching the Fourier calculation.
\[ \boxed{\, \dfrac{\pi^{2}}{8} \,} \]