Using Fourier series to evaluate an infinite odd series

Expert intro: This problem combines half-range Fourier series, symmetry, and series evaluation. The key is to build the correct odd extension, compute the coefficients carefully, and then plug in a convenient value of $x$ to extract the target sum.

Detailed walkthrough

Identify the symmetry and the right extension

The first step is to use the given function on $0\le x<\pi$ and extend it to $-\pi\le x<0$ in a way that matches the Fourier strategy. For an odd extension, the function must satisfy $f(-x)=-f(x)$. That means the negative-side values are not chosen arbitrarily; they must mirror the positive-side values with a sign change.

Once the odd periodic extension is built, the Fourier series will contain only sine terms. That is the main simplification: odd functions have no cosine terms and no constant term in their Fourier expansions.

Compute the Fourier coefficients

For an odd $2\pi$-periodic function, the sine coefficients are

$$b_n=\frac{2}{\pi}\int_0^\pi f(x)\sin(nx)\,dx.$$

Using the given expression on $[0,\pi)$, you evaluate the integral with integration by parts, and the result collapses to a term involving $1-(-1)^n$. This factor is important because it shows that only odd values of $n$ survive.

That means the series can be rewritten with only odd harmonics. In practice, this turns a full infinite sum into a much cleaner odd-indexed series, which is exactly what makes the final evaluation manageable.

Write the simplified Fourier series

Because $1-(-1)^n=0$ when $n$ is even and $2$ when $n$ is odd, the series may be written using only odd $n$ values. The Fourier expansion therefore has the form

$$f_{odd}(x)=\sum_{k=1}^{\infty} c_{2k-1}\sin((2k-1)x),$$

with coefficients determined from the integral result. The simplification is not cosmetic; it is the reason the final series can be matched to a known trigonometric identity at a specific point.

Now evaluate the series at $x=\frac{\pi}{2}$. Since

$$\sin\left((2k-1)\frac{\pi}{2}\right)=(-1)^{k-1},$$

the series becomes an alternating sum over odd denominators. That is the bridge from Fourier theory to the infinite series value.

Extract the infinite series value

Substituting $x=\frac{\pi}{2}$ into the simplified Fourier series gives the desired odd-square series. The resulting identity is the standard form

$$1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots$$

or the equivalent alternating odd reciprocal-square sum, depending on the exact coefficient normalization used in your setup.

The important method is the same in every version of this problem:

1. build the correct odd extension,
2. compute the Fourier coefficients,
3. remove zero terms using parity,
4. evaluate at a special point such as $x=\pi/2$.

That sequence converts a Fourier series into a closed-form infinite series identity.

💡 Pitfall guide

A common mistake is extending the function as even or copying the original formula onto the negative interval without enforcing odd symmetry. That changes the entire Fourier basis and produces cosine terms that do not belong. Another frequent error is forgetting that $1-(-1)^n$ kills every even index, so keeping even terms in the final sum leads to the wrong series. Finally, when evaluating at $x=\pi/2$, do not simplify the sine values too early; keep the parity pattern clear so the alternating sign is tracked correctly.

🔄 Real-world variant

If the function were extended as an even function instead of an odd one, the Fourier series would switch from sine terms to cosine terms. For example, a variant problem might ask: "Let $f(x)=x^2$ on $0\le x<\pi$ and extend it evenly to $[-\pi,\pi]$. Find the cosine series and use it to evaluate a related sum." In that case, the coefficient formula changes, the symmetry changes, and the target infinite series would typically involve non-alternating reciprocal squares rather than an alternating odd-indexed sum.

🔍 Related terms

half-range Fourier series, odd periodic extension, Parseval identity