limit of supremum of a sequence Can any one help me with this? 
Let $c$ be a real number. I would like to show that
$$ \limsup_{n \to \infty}\sqrt[n]{\left|\frac{i}{2}\left(\frac{(c-i)^{n+1}-(c+i)^{n+1}}{c^{2}+1}\right)\right|}=\sqrt{c^{2}+1}.$$
I came up with this when I was trying to prove that the radius of convergence for the power series of then function $\frac{1}{x^2+1}$ at a real point $c$ is $\sqrt{c^2+1}$.
 A: Write $c+\mathrm i=r\mathrm e^{\mathrm it}$ with $r=\sqrt{c^2+1}$ and $t$ in $(0,\pi)$ such that $\cos(t)=c/r$. Then, the LHS is
$$
R_n=\sqrt[n]{\left|r^{n-1}\,\sin((n+1)t)\right|}=r^{1-1/n}\cdot u_n,
$$
with
$$
u_n=\sqrt[n]{\left|\sin((n+1)t)\right|}.
$$
Then $r^{1-1/n}\to r$. Note that there is no reason to believe the sequence $(u_n)$ should converge. For example, if $c=1$, $t=\pi/4$, hence it has two limit points: $0$ for the subsequence $(u_{4n+3})$ (which is identically $0$), and $1$ otherwise. 
Fortunately, the question asks for the limsup of $(R_n)$, not for its (nonexistent) limit. The key remark here is that, for every fixed $t$ in $(0,\pi)$, $u_n^n=\left|\sin((n+1)t)\right|$ is periodically at least some $\varepsilon(t)\gt0$. On the other hand, $u_n\leqslant1$ for every $n$, hence
$$
1=\lim\limits_{n\to\infty}\sqrt[n]{\varepsilon(t)}\leqslant\limsup\limits_{n\to\infty}\ u_n\leqslant1.
$$
Here is a proof that $\varepsilon(t)$ exists. Since $t\gt0$, the sequence of angles $(n+1)t$ grows to infinity, and since $t\lt\pi$, it jumps strictly less than $\pi$ at a time, hence it must land periodically in an interval of length $t$ centered at a point of $\pi/2+2\pi\mathbb Z$. On such an interval, the sine is uniformly at least $\varepsilon(t)=\cos(t/2)\gt0$, QED.
Finally,
$$
\limsup\limits_{n\to\infty}\ R_n=r\cdot\limsup\limits_{n\to\infty}\ u_n=r=\sqrt{c^2+1}.
$$
