In their foundational paper "Vector bundles and homogeneous spaces," Atiyah and Hirzebruch show, among many other things, that for $G$ a compact, connected Lie group, the K-theory of the classifying space $BG$, taken as an inverse limit of the K-theory of a family of compact approximations of $B_n G$, is isomorphic to the completion of the representation ring $R(G)$ at the augmentation ideal $I(G)$:
$$K^*(BG) := \varprojlim K^*(B_nG) \cong \varprojlim R(G)/I(G)^n =: \widehat R(G).$$
They make a point of noting that the Chern character, which is an isomorphism upon rationalizing on every compact approximation, extends to a map from $K^*(BG)$ to the completion of $H^*(BG;\mathbb Q)$ at the augmentation ideal $H^{\geq 1}(BG;\mathbb Q)$. They do not state that this becomes an isomorphism on tensoring with $\mathbb Q$, and I have seen it asserted that in fact it is not.
Why not?
Here is an argument that in fact it should be so for $G = S^1 = \mathrm{U}(1)$; perhaps someone can tell me where I've gone wrong.
The cohomology is $H^*(BS^1) = \mathbb{Z}[c]$, where $c \in H^2(BS^1)$ is the first Chern class. The completion is clearly $\hat H^*(BS^1;\mathbb Q) = \mathbb Q[[c]]$.
On the other hand, the representation ring is $R(S^1) = \mathbb Z[s,s^{-1}]$, where $s$ is the standard representation as scalar multiplication by the unit circle in $\mathbb C$. Atiyah–Hirzebruch argue the completion at the augmentation ideal $I(S^1) = (s-1) = (s-1,s^{-1}-1)$ is $\widehat R(S^1) = \mathbb Z[[s-1]]$. Tensoring with the rationals gives $\mathbb Q[[s-1]]$ [edit: This is untrue; see Eric Wofsey's answer.].
Now the Chern character takes $s \mapsto e^c$ in each approximation $H^*(B_n G)$, so it should take $$s-1 \longmapsto e^c -1 = c + c^2/2 + \cdots \in \hat H^*(BS^1).$$
It seems clear this is injective. As for surjectivity, as there is no constant term in $f(c) = e^c - 1$, it admits a composition-inverse $g$ in $\mathbb Q[[c]]$, and then
$$g(s-1) \longmapsto g(f(c)) = c.$$
A similar argument would establish the same for $G = T$ a torus. What goes wrong?
The problem is that it is not true that the canonical map $i:\mathbb{Z}[[x]]\otimes\mathbb{Q}\to\mathbb{Q}[[x]]$ is an isomorphism. Any power series of the image of $i$ must have only finitely many different denominators on its coefficients, since an element of $\mathbb{Z}[[x]]\otimes\mathbb{Q}$ is a sum involving only finitely many elements of $\mathbb{Q}$. So for instance, the power series $\sum x^n/n!$ is not in the image of $i$.
This means that your argument breaks down, and actually the completion of $H^*(BS^1;\mathbb{Q})$ is much larger than the image of $\widehat{R}(S^1)\otimes\mathbb{Q}$ under the Chern character (specifically, your argument shows that the Chern character extends to an isomorphism $\mathbb{Q}[[s-1]]\to \mathbb{Q}[[c]]$, and the image of the Chern character is just the image of $\mathbb{Z}[[s-1]]\otimes\mathbb{Q}$ under this isomorphism).