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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?

Scientifica
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clathratus
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    As an alternative to solve or to check the result (without using wolfy) recall that by telescoping $$\sum_{n\geq1}\frac1{n(n+1)}=\sum_{n\geq1}\frac1{n}-\sum_{n\geq1}\frac1{n+1}=1\color{red}{-\frac12+\frac12-\frac13+\frac13-\frac14+\ldots}=1$$ – user Nov 02 '18 at 17:38

1 Answers1

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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.

Siong Thye Goh
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    To make this clearer I would say the error comes earlier, when you relabel $\tfrac{1}{n}x^n$ as the indefinite integral of $x^{n-1}$. It should be $\int_0^x t^{n-1}dt$. – SBK Nov 02 '18 at 16:34