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\[\frac{ .4 }{ \pi }\sum_{n=1}^{\infty} \frac{ 1 }{ n }\sin(\frac{ n*\pi }{ 5 })\cos(\frac{n*\pi*t}{2})=....\]
\[\frac{ .4 }{ \pi }\sum_{n=1}^{\infty} \frac{ 1 }{ n }\sin(\frac{ n*\pi }{ 5 })\sum_{n=1}^{\infty} \frac{ 1 }{ n }\cos(\frac{n*\pi*t}{2})\]
I don't think so, \[\sum_{n=1}^\infty a_nb_n = a_1b_1 + a_2b_2+a_3b_3+\cdots\]\[\sum_{n=1}^\infty a_n\sum_{n=1}^\infty b_n = (a_1+a_2+a_3+\cdots)(b_1+b_2+b_3+\cdots)\]
Right? @optiquest
Yes
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