Recall (Binomial Series)
(1+x)^\alpha=\sum_{n=0}^\infty\frac{\alpha\cdot(\alpha-1)\cdot(\alpha-2)\cdot\cdots\cdot(\alpha-n+1)}{n!}x^n(1+x)α=∞∑n=0α⋅(α−1)⋅(α−2)⋅⋯⋅(α−n+1)n!xn
Let us look at f(x)f(x).
f(x)=\frac{1}{(1+x)^2}=(1+x)^{-2}f(x)=1(1+x)2=(1+x)−2
By Binomial Series with \alpha=-2α=−2,
=\sum_{n=0}^\infty\frac{(-2)\cdot(-3)\cdot(-4)\cdot\cdots\cdot(-n)\cdot(-(n+1))}{n!}x^n=∞∑n=0(−2)⋅(−3)⋅(−4)⋅⋯⋅(−n)⋅(−(n+1))n!xn
By factoring out the negative signs,
=\sum_{n=0}^\infty(-1)^n\frac{\cancel{2}\cdot\cancel{3}\cdot\cancel{4}\cdot\cdots\cdot \cancel{n}\cdot(n+1)}{1\cdot\cancel{2}\cdot\cancel{3}\cdot\cancel{4}\cdot\cdots\cdot \cancel{n}}x^n
By cleaning up,
=\sum_{n=0}^\infty(-1)^n(n+1)x^n
I hope that this was clear.
