Prove that for -1<x<1 1/(x+1) = infinite Σ [(-1)^n][x^n] = 1 - x + x^2 - x^3 + ... n=0 ln(1+x) = infinite Σ [(x^n)(-1)^(n-1)]/n = x - (x^2)/2 + (x^3)/3 - (x^4)/4... n=0
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thats a geo $\displaystyle {1\over 1+x}={1\over 1-(-x)}=\sum_{n=0}^{\infty}(-x)^n$ Next integrate AND do not forget the +C, which by setting x=0 becomes 0. AND your sum is wrong, you must start at n=1. You cannot divide by 0.
thank you!
Originally Posted by wyhwang7 Prove that for -1<x<1 infinite Σ [(x^n)(-1)^(n-1)]/n = x - (x^2)/2 + (x^3)/3 - (x^4)/4... n=0 I hope your book does not start this sum at zero. It looks like 1 to me. You cannot divide by 0, anywhere.
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