An Elementary Proof for Euler's Series

Daniel J. Velleman
Am Math Monthly, V 123, N 1, Jan 2016, p. 77

In an earlier note, Yoshio Matsuoka gave an elementary proof for the sum of Euler's series, $\displaystyle\sum_{i=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.\;$ Below we sketch a simplified version of Matsuoka's proof.

For every integer $n\ge 0,\;$ let $\displaystyle I_n=\int_{0}^{\frac{\pi}{2}}\cos^{2n}xdx\;$ and $\displaystyle J_n=\int_{0}^{\frac{\pi}{2}}x^2\cos^{2n}xdx.\;$ Clearly, $\displaystyle I_0=\frac{\pi}{2}\;$ and $\displaystyle J_0=\frac{\pi^3}{24}.\;$ For $n\ge 1,\;$ we evaluate $I_n\;$ using integration by parts, with $dv=\cos xdx;\;$ we get $I_n=(2n-1)(I_{n-1}-I_n), and subsequently,

(1)

$\displaystyle I_n=\frac{2n-1}{2n}\cdot I_{n-1}.$

Alternatively, we can apply integration by parts twice, first with $dv=\cos xdx\;$ and then with $dv=2xdx,\;$ to obtain

(2)

$I_n=n(2n-1)J_{n-1}-2n^2J_n.$

Dividing (2) by $n^2I_n\;$ and then applying (1), we find that

(3)

$\displaystyle \frac{1}{n^2}=2\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right).$

We now sum (3) for $n\;$ from $1\;$ to $N,\;$ and note that the right-hand side telescopes:

(4)

$\displaystyle \frac{1}{n^2}=2\left(\frac{J_{0}}{I_{0}}-\frac{J_N}{I_N}\right)=\frac{\pi^2}{6}-2\cdot\frac{J_N}{I_N}.$

Finally, we use the inequality $\displaystyle x\le\frac{\pi}{2}\sin x,\;$ for $\displaystyle 0\le x\le\frac{\pi}{2}\;$ and (1) to estimate $J_N\;$ as follows:

$\displaystyle\begin{align} 0\le J_N &\le\frac{\pi^2}{4}\int_{0}^{\frac{\pi}{2}}\sin^2x\ \cos^{2n}xdx\\ &=\frac{\pi^2}{4}\left(I_N-I_{N+1}\right)\\ &=\frac{\pi^2}{4}\cdot\frac{1}{2N+2}\cdot I_N. \end{align}$

Therefore, $\displaystyle\lim_{N\rightarrow\infty}\frac{J_N}{I_N}=0\;$ Letting $N\rightarrow\infty\;$ in (4), the conclusion follows.

References

  1. Y. Matsuoka, An elementary proof of the formula $\sum_{n=1}^{\infty}1/n^2=\pi^2/6$, Amer. Math. Monthly 68 no. 5 (1961) 485487, https://www.jstor.org/stable/2311110.

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