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. 
 A: $$x - x^{1/2} + x^{1/3} - \ldots = (-1)^{1+ 1} (x)^{1/1} + (-1)^{2+1} (x)^{1/2} + \ldots = \sum_{n=1}^{\infty} (-1)^{n+1} x^{1/n}$$
This series doesn't converge for $0 < x \le 1$. We can't integrate it.
Edit:
For $0 < x \le 1$:
$$\lim_{n \to \infty} \sup \{ (-1)^{n+1} x^{1/n} \}= 1 \neq 0$$
The series fails to converge (using the limit test).
A: $$\int_0^1 x^{1/n}dx=\frac{n}{n+1}$$ hence $$K=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n+1}$$ 
you should be able to continue ...
A: If $x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\cdots$ converges then $\lim\limits_{n\to\infty} \sqrt[n] x =0$.  But in fact $\lim\limits_{n\to\infty} \sqrt[n] x =1$, so the function to be integrated is not well defined in the usual sense.
An integral $\displaystyle \int(A+B+C+\cdots)$ can be written as $\displaystyle\int A + \int B+\int C+\cdots$ if there are only finitely many terms, but it doesn't always work with infinitely many terms, and one might wonder whether a convergent series would result in this case:
$$
\int_0^1 (x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\cdots)\, dx = \frac 1 2 - \frac 2 3 + \frac 3 4 - \frac 4 5 + \cdots.
$$
But again, the absolute values of the terms approach $1$ rather than $0$, so the series diverges.
There are non-standard "summation methods" by which series that don't converge in the simple sense taught in high school have sums in another sense.  Whether any of those can be applied here I don't know.
A: 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
$$
\int_0^1 ( x - \sqrt{x} + \sqrt[3]{x} - \sqrt[4]{x} + \cdots ) = \ln(2) - 1
$$
A: However, if we interpret $x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\ldots$ as $\sum_{n=1}^\infty\,\left(x^{\frac{1}{2n-1}}-x^{\frac{1}{2n}}\right)$, then this series converges uniformly on any interval $[\epsilon,1]$ for $\epsilon>0$.  Hence, the required integral would then be $\sum_{n=1}^\infty\,\left(\frac{2n-1}{2n}-\frac{2n}{2n+1}\right)=\sum_{n=1}^\infty\,\left(-\frac{1}{2n}+\frac{1}{2n+1}\right)=-1+\ln(2)$.
