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 $$ u=1+\frac{a}{|x|^{n-2}}+O(|x|^{1-n})\quad ? $$
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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.
