TITLE: The expansion of harmonic function at infinity
QUESTION [5 upvotes]: If $u$ is a harmonic function on $\mathbb R^n$ outside some compact set such that $u$ goes to $1$ at infinity. Then does $u$ have the following expansion
$$
u=1+\frac{a}{|x|^{n-2}}+O(|x|^{1-n})\quad ?
$$

REPLY [6 votes]: Yes, this is true (in dimensions $n>2$).    Let's subtract $1$ from $u$, so that $u\to 0$ at infinity. Let $v$ be the Kelvin transform of $u$, that is 
$$
v(x) = \|x\|^{2-n}u(x/\|x\|^{2})
$$
Since $u$ is harmonic outside of a compact set, $v$ is harmonic in a punctured neighborhood of $0$. The singularity at $0$ is removable since $v(x)=o(\|x\|^{2-n})$ as $x\to 0$. (Sketch of proof: $v(x) \pm \epsilon \|x\|^{2-n}$ is sub/super-harmonic; write down the corresponding sub/super-mean value properties and  let $\epsilon\to 0$ to get the  mean value property for $v$.)  
Being harmonic in a neighborhood of $0$, $v$ is analytic there: $v(x) = \sum_{k=0}^\infty p_k(x)$ where  $p_k$ is a homogeneous polynomial of degree $k$. Return to $u$: 
$$
u(x) = \|x\|^{2-n}v(x/\|x\|^{2})=\sum_{k=0}^\infty \|x\|^{2-n}p_k(x/\|x\|^{2})
 = \sum_{k=0}^\infty \|x\|^{2(1-k)-n}p_k(x)
$$
which yields the desired asymptotics.