# Assuming the expansion for $(1+x)^{-1}$ prove that $\int_0^1 \frac{x \mathrm d x}{1+x}=\frac12-\frac13+\frac14-\frac15+\cdots$

Given: $(1+x)^{-1}=1-x+x^2-x^3+\cdots$ for $-1<x<1$, prove that

$\int_0^1 \frac{x \mathrm d x}{1+x}=\frac12-\frac13+\frac14-\frac15+\cdots$

My attempt: I multiplied both sides of $(1+x)^{-1}=1-x+x^2-x^3+\cdots$ for $-1<x<1$ by $x$ to get:

$\frac{x}{1+x}=x-x^2+x^3-x^4+\cdots$ for $-1<x<1$ and then I integrate term by term within $(0,1)$ to get the result.

But my question is, is such a method valid? The multiplication part I mean. If so, why? I have been able to do three similar sums by this method, but I do not understand the reasoning.

• If you had waited a little more before accepting this, you would have known... – Did Nov 14 '14 at 21:27
• Term-by-term integration of a power series certainly holds on the interior of the interval of convergence. But here we integrate up to the boundary point $1$, where the series $(1+x)^{-1}$ diverges. So it takes a bit of thought to complete. – GEdgar Nov 14 '14 at 21:59

Yes, this is a valid operation. Formally, it is always a valid thing to do, but to show that the sum actually still converges to the right number, you can reexamine the remainder of the finite sum. Operations like this are always valid inside the radius of convergence, though sometimes they will move the interval of convergence (for example, substituting $z-4$ for $x$ changes the interval of convergences to $(3,6)$)