# 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 ### Related materialRead more... ### Telescoping situations • Leibniz and Pascal Triangles • Infinite Sums and Products • Sum of an infinite series • Harmonic Series And Its Parts • A Telescoping Series • An Inequality With an Infinite Series • That Divergent Harmonic Series •$\sin 1^{\circ}+\sin {2^\circ}+\sin 3^{\circ}+\cdots+\sin 180^{\circ}=\tan 89.5^{\circ}$• Problem 3824 from Crux Mathematicorum •$x_n=\sin 1+\sin 3+\sin 5+\cdots+\sin (2n-1)\$
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