What part of this is this wrong? $\sum_{n\geq1}\frac1{n(n+1)}=-1$

Question

Here is my attempt to find $$P=\sum_{n\geq1}\frac1{n(n+1)}$$ First we note that $$\int_0^1 x^n dx=\frac1{n+1}$$ Plugging in gives $$P=\sum_{n\geq1}\frac1n\int_0^1x^ndx$$ $$P=\int_0^1\sum_{n\geq1}\frac1nx^ndx$$ $$P=\int_0^1\sum_{n\geq1}\int x^{n-1}dxdx$$ $$P=\int_0^1 \int\sum_{n\geq1}x^{n-1}dxdx$$ $$P=\int_0^1 \int\sum_{n\geq0}x^{n}dxdx$$ $$P=\int_0^1 \int\frac{dx}{1-x}dx$$ $$P=\int_0^1 \log(1-x)dx$$ $$P=-1$$ But Wolfram alpha says $P=1$. What did I do wrong?

Answer 1

Note that

$$\int_0^x \frac1{1-t}\, dt = -\ln(1-t)|_{t=0}^{t=x}=-\ln(1-x)$$

Side remark:

You have reuse $x$ for different purposes. Try to avoid that.