Evaluate:
∞∑j=1∞∑i=1j2i3j(j3i+i3j)
Honestly, I don't see where to start with this. I am sure that this is a trick question and I am missing something very obvious. I tried writing down a few terms for a fixed j but I couldn't spot any pattern or some kind of easier series to handle.
Any help is appreciated. Thanks!
Answer
After symmetrization with respect to the exchange i↔j, the sum can be rewritten as
12∞∑i,j=1(j2i3j(j3i+i3j)+i2j3i(j3i+i3j))=12∞∑i,j=1i⋅j3i⋅3j=12(∞∑i=1i3i)2=932.
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