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Complex series

Can ∑n=0∞in3\sum_{n=0}^\infty i n^3 be written as i∑n=0∞n3i\sum_{n=0}^\infty n^3, where i i is the imaginary unit?
Neither sum converges...
Reply 2
Avatar for John taylor
John taylor
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1
Original post by IrrationalNumber
Neither sum converges...


Probably, but i just made them up. Just want to know if i can take the imaginary unit ii out.
Original post by John taylor
Probably, but i just made them up. Just want to know if i can take the imaginary unit ii out.


It's no different to any other constant in that regard.
Reply 4
Avatar for natninja
natninja
21
Original post by John taylor
Can ∑n=0∞in3\sum_{n=0}^\infty i n^3 be written as i∑n=0∞n3i\sum_{n=0}^\infty n^3, where i i is the imaginary unit?


∑n=0∞in3\sum_{n=0}^\infty i n^3

I think may be written as

∑n=0∞i∑n=0∞n3\sum_{n=0}^\infty i \sum_{n=0}^\infty n^3
(edited 10 years ago)
Reply 5
Avatar for John taylor
John taylor
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Original post by natninja
∑n=0∞in3\sum_{n=0}^\infty i n^3

I think may be written as

∑n=0∞i∑n=0∞n3\sum_{n=0}^\infty i \sum_{n=0}^\infty n^3


If you can treat i i as a constant, why do you have a summation in front of it?
Reply 6
Avatar for natninja
natninja
21
Original post by John taylor
If you can treat i i as a constant, why do you have a summation in front of it?


if you do the summation of in^3 then you get

i+8i+27i+64i+125i etc.

so... apparently I'm wrong...
Reply 7
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DFranklin
18
Original post by natninja
∑n=0∞in3\sum_{n=0}^\infty i n^3

I think may be written as

∑n=0∞i∑n=0∞n3\sum_{n=0}^\infty i \sum_{n=0}^\infty n^3
No it can't.

As others have suggested:

∑0∞ian=i∑0∞an\sum_0^\infty i a_n = i \sum_0^\infty a_n
Reply 8
Avatar for natninja
natninja
21
Original post by DFranklin
No it can't.

As others have suggested:

∑0∞ian=i∑0∞an\sum_0^\infty i a_n = i \sum_0^\infty a_n


I realised see above ^

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