If |x|<1, Then the sum of the series 2x−11−x+x2+4x3−2x1−x2+x4+8x7−4x31−x4+x8+⋯⋯
Try: Let A=1−2x1−x+x2+4x3−2x1−x2+x4+8x7−4x31−x4+x8+⋯
∫Adx
=∫[1−2x1−x+x2+4x3−2x1−x2+x4+8x7−4x31−x4+x8+⋯⋯]dx
∫Adx=ln[(1−x+x2)⋅(1−x2+x4)⋅(1−x4+x8)⋯]
Now i have seems that expression under ln on
Right side must have closed form in $-1
But i could not understand how can i find it.
could some help me, thanks
Answer
Hint:
(1+x+x2)(1−x+x2)=(1+x2)2−x2=?
⟹n∏r=1(1−x2r+(x2r)2)=1−x2n+1+(x2n+1)21+x+x2
Now limn→∞2n+1→∞
and |x|<1,limm→∞xm=0
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