+0

# 2016 NS 25

0
202
5
+935

Can anyone solve this?

Is there a formula for this?

Mar 13, 2019

#1
+6196
+2

$$\dfrac{\frac 7 8 \zeta(3)}{\zeta(3)} = \dfrac 7 8$$

.
Mar 13, 2019
#2
+935
+1

How did you get that? And what is that notation?

dgfgrafgdfge111  Mar 14, 2019
#3
+6196
+2

That came out of Mathematica.

The squiggle stands for the Riemann Zeta function.

It's really more advanced stuff than I know.

Rom  Mar 14, 2019
#4
+6196
+2

$$\text{I have a simpler answer.}\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac{1}{(2k)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1$$

$$\sum \limits_{k=1}^\infty \dfrac{1}{(2k)^3} = \sum \limits_{k=1}^\infty \dfrac{1}{8k^3} = \dfrac 1 8 \sum \limits_{k=1}^\infty \dfrac{1}{k^3}$$

$$\dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}+\dfrac 1 8 \sum \limits_{k=1}^\infty\dfrac{1}{k^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}}=1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} + \dfrac 1 8 = 1\\ \dfrac{\sum \limits_{k=1}^\infty~\dfrac{1}{(2k-1)^3}}{\sum \limits_{k=1}^\infty\dfrac{1}{k^3}} = \dfrac 7 8$$

.
Mar 14, 2019
edited by Rom  Mar 14, 2019
#5
+935
0

Thanks for the help!!!

dgfgrafgdfge111  Mar 14, 2019