Integral calculation: $\int_0^1 (x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\cdots) \, dx$ $$K=\int_0^1 (x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\cdots) \, dx$$
I'm looking for an exact solution for this integral. Thank you very much. 

Given
$$
K = \int_0^1 ( x - \sqrt{x} + \sqrt[3]{x} - \sqrt[4]{x} + \cdots )
$$
Write this as
$$
K = - \int_0^1 \sum_{k=1}^\infty (-1)^k x^{1/k}
$$
Thus
$$
K = - \sum_{k=1}^\infty (-1)^k \int_0^1 x^{1/k} = \sum_{k=1}^\infty (-1)^k \left[ \frac{1}{1/k+1} x^{1/k+1} \right]_0^1
$$
So
$$
K = - \sum_{k=1}^\infty (-1)^k \frac{k}{1+k}
$$
We obtain
$$
K = \frac{1}{2} - \frac{2}{3} + \frac{3}{4} - \frac{4}{5} + \cdots
$$
or
$$
K = - \frac{1}{2 \cdot 3} - \frac{1}{4 \cdot 5} - \frac{1}{6 \cdot 7} - \cdots
$$
or
$$
K = - \frac{1}{6} - \frac{1}{20} - \frac{1}{42} - \cdots
$$
Which can be written as $\ln(2) - 1$, so
$$
\int_0^1 ( x - \sqrt{x} + \sqrt[3]{x} - \sqrt[4]{x} + \cdots ) = \ln(2) - 1
$$
