If we evaluate the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi,$ we get

$$\begin{equation*}x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\tag{1}\end{equation*}$$

So, for $f(\pi )=\pi ^{2p}$ we have

$$\begin{equation*}\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\tag{2}\end{equation*}$$

where the integral

$$\begin{equation*}I_{2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x\tag{3}\end{equation*}$$

satisfies the following recurrence, as can be shown by integration by parts

$$\begin{equation*}I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}I_{2\left( p-1\right) },\qquad I_{0}=0.\tag{4}\end{equation*}$$

• For $p=1$, we obtain $$\begin{equation*}I_{2}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi.\end{equation*}\tag{5}$$ and

$$\pi ^{2}=\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\left(\frac{2}{n^{2}}\pi \cos n\pi \right)=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}}.\tag{6}$$ Consequently, $$\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}\tag{7}$$

By using $(1)$ to $(4)$ we can generate recursively the sequence $(\zeta(2p))_{p\ge 1}$. For instance, I evaluated $\zeta(4)$ and $\zeta(6)$ in this math.SE answer.

If we evaluate the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi,$ we get

$$\begin{equation*}x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\tag{1}\end{equation*}$$

So, for $f(\pi )=\pi ^{2p}$ we have

$$\begin{equation*}\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\tag{2}\end{equation*}$$

where the integral

$$\begin{equation*}I_{2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x\tag{3}\end{equation*}$$

satisfies the following recurrence, as can be shown by integration by parts

$$\begin{equation*}I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}I_{2\left( p-1\right) },\qquad I_{0}=0.\tag{4}\end{equation*}$$

• For $p=1$, we obtain $$\begin{equation*}I_{2}=\frac{2}{n^{2}}\pi \cos n\pi.\end{equation*}\tag{5}$$ and

$$\pi ^{2}=\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\left(\frac{2}{n^{2}}\pi \cos n\pi \right)=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}}.\tag{6}$$ Consequently, $$\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}\tag{7}$$

By using $(1)$ to $(4)$ we can generate recursively the sequence $(\zeta(2p))_{p\ge 1}$. For instance, I evaluated $\zeta(4)$ and $\zeta(6)$ in this math.SE answer.

If we evaluate the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi,$ we get

$$\begin{equation*}x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\tag{1}\end{equation*}$$

So, for $f(\pi )=\pi ^{2p}$ we have

$$\begin{equation*}\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\tag{2}\end{equation*}$$

where the integral

$$\begin{equation*}I_{2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x\tag{3}\end{equation*}$$

satisfies the following recurrence, as can be shown by integration by parts

$$\begin{equation*}I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}I_{2\left( p-1\right) },\qquad I_{0}=0.\tag{4}\end{equation*}$$

• For $p=1$, we obtain $$\begin{equation*}I_{2}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi.\end{equation*}\tag{5}$$ and

$$\pi ^{2}=\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\left(\frac{2}{n^{2}}\pi \cos n\pi \right)=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}}.\tag{6}$$ Consequently, $$\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}\tag{7}$$

By using $(1)$ to $(4)$ we can generate recursively the sequence $(\zeta(2p))_{p\ge 1}$. For instance, I evaluated $\zeta(4)$ and $\zeta(6)$ in this math.SE answer.

If we evaluate the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi,$ we get

$$\begin{equation*}x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\tag{1}\end{equation*}$$

So, for $f(\pi )=\pi ^{2p}$ we have

$$\begin{equation*}\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\tag{2}\end{equation*}$$

where the integral

$$\begin{equation*}I_{2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x\tag{3}\end{equation*}$$

satisfies the following recurrence, as can be shown by integration by parts

$$\begin{equation*}I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}I_{2\left( p-1\right) },\qquad I_{0}=0.\tag{4}\end{equation*}$$

• For $p=1$, we obtain $$\begin{equation*}I_{2}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi.\end{equation*}\tag{5}$$ and

$$\pi ^{2}=\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\left(\frac{2}{n^{2}}\pi \cos n\pi \right)=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}}.\tag{6}$$ Consequently, $$\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}\tag{7}$$

By using $(1)$ to $(4)$ we can generate recursively the sequence $(\zeta(2p))_{p\ge 1}$. For instance, I evaluated $\zeta(4)$ and $\zeta(6)$ in this math.SE answer.