1+19+145+1189+1729+⋯=∞∑n=11(2n−1)⋅3n−1
I got:∞∑n=11(2n−1)⋅3n−1=∞∑n=11∫0x2n−2dx3n−1=…
And no idea how to impove.
Thx!
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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