Are you familiar with this series?

Calculus Level 3

$\large 1 - \dfrac{1}{4} + \dfrac{1}{9} - \dfrac{1}{16} + \cdots = \, ?$

None of these

\dfrac{\pi^2}{6}

\dfrac{\pi^2}{8}

\dfrac{\pi^2}{4}

\ln 2

\dfrac{\pi^2}{12}

\dfrac{\pi^2}{24}

2 solutions

Akshat Sharda
Sep 29, 2015

$1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \ldots \infty$

$1+\frac{1}{4}+\frac{1}{9}+\ldots \infty -2\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\ldots \infty \right)$

$1+\frac{1}{4}+\frac{1}{9}+\ldots \infty-\frac{1}{2}\left(1+\frac{1}{4}+\frac{1}{9}+\ldots \infty \right)$

We know ,

$1+\frac{1}{4}+\frac{1}{9}+\ldots \infty=\frac{\pi^{2}}{6}$

For proofs see here .

$\frac{\pi^{2}}{6}-\frac{1}{2}\cdot \frac{\pi^{2}}{6}=\frac{\pi^{2}}{6}-\frac{\pi^{2}}{12}=\boxed{\frac{\pi^{2}}{12}}$

Tom Kovar
Mar 5, 2017