Johnny: Difference between revisions

Latest revision as of 16:03, 23 August 2022

$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{\frac {m^{3}2n}{4^{m}\left(m6^{n}+n8^{m}\right)}}$

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	<math>\sum_{n=1}^\infty\{n^2}{3n}		<math>\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^3 2n}{4^m\left(m 6^n + n 8^m\right)}</math>