Can we use that?

Algebra Level 4

It is given that $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^{4}} = \dfrac{\pi^{4}}{90}$ and the value of $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{(2n-1)^{4}}$ can be represented as $\dfrac{\pi^{a}}{b}$ , find $a+b$ .

The answer is 100.

1 solution

Chew-Seong Cheong
Mar 27, 2016

$\begin{aligned} S & = \sum_{n=1}^\infty \frac{1}{(2n-1)^4} \\ & = \sum_{n=1}^\infty \frac{1}{n^4} - \sum_{n=1}^\infty \frac{1}{(2n)^4} \\ & = \sum_{n=1}^\infty \frac{1}{n^4} - \frac{1}{2^4} \sum_{n=1}^\infty \frac{1}{n^4} \\ & = \left(1-\frac{1}{16} \right) \color{#3D99F6}{\sum_{n=1}^\infty \frac{1}{n^4} \quad \quad \small \text{Riemann zeta function } \zeta (4) = \sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}} \\ & = \frac{15}{16} \times \color{#3D99F6}{\frac{\pi^4}{90}} \\ & = \frac{\pi^4}{96} \end{aligned}$

$\Rightarrow a + b = 4+96 = \boxed{100}$

More about Riemann zeta function .

Did it same.

Dev Sharma - 5 years, 2 months ago

Problem woluld get less solvers if the first information is not given

Aakash Khandelwal - 5 years, 2 months ago

It is actually Riemann zeta function. I should mention it so that those who don't know will learn.

Chew-Seong Cheong - 5 years, 2 months ago

Can we use that?

The answer is 100.

1 solution

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