Pie, anyone?

Calculus Level 3

$\begin{aligned}\large 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\ldots &=& \large \frac{π^2}{6} \\ \large 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\ldots &=& \large \ ? \end{aligned}$

\frac{π^{2}}{10}

\frac{π^{2}}{9}

\frac{π^{2}}{8}

\frac{π^{2}}{12}

2 solutions

Brian Charlesworth
May 4, 2015

The equation on the first line can be written as $S = \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k^{2}} = \dfrac{\pi^{2}}{6}.$

The equation on the second line can then be written as

$S - \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{(2k)^{2}} = S - \dfrac{1}{4}\sum_{k=1}^{\infty}\dfrac{1}{k^{2}} = S - \dfrac{1}{4}S = \dfrac{3}{4}*\dfrac{\pi^{2}}{6} = \boxed{\dfrac{\pi^{2}}{8}}.$

Thanks for the solutions...

Kun Yi Ong - 6 years, 1 month ago

Simply superb.

Pushparaju Muddarusu - 6 years, 1 month ago

But according to you answer should not be D

Abhishek Gupta - 5 years, 10 months ago

I'm sorry, but I'm not quite clear on what your concern is. $\dfrac{\pi^{2}}{8}$ is one of the answer options and is the correct option, just as I have in my solution.

Brian Charlesworth - 5 years, 10 months ago

Yinjoe Ng
Jan 13, 2021

Multiply S=1/4+1/16+1/36+1/64+......by 4 and we can get that 4S=pi^2/6 so S=pi^2/24. Subtract pi^2/6 by S and we get the sum for the required series.

Pie, anyone?

2 solutions

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