# 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 $\sqrt{c^2+1}$.

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I edited your post. I hope that I maintained your original question, but please edit the post if I accidentally changed the meaning of anything. Also, note that $|i| = 1$ and $s^{1/n} \to 1$ for any positive number $s$. Thus, you need only consider $$\limsup_{n \to \infty} \sqrt{|(c-i)^{n+1} - (c+i)^{n+1}|}.$$ – JavaMan Jan 19 '12 at 19:46
@JavaMan You missed an "n" in the radical sign in your comment. – Srivatsan Jan 19 '12 at 19:48
@Srivatsan: Thanks for pointing that out. Unfortunately I am only now seeing your comment, and I am past the 5 minute window where I can edit the comment... – JavaMan Jan 19 '12 at 19:56
If the question above refers to this answer, you could have mentioned the fact. – Did Jan 19 '12 at 20:16
@JavaMan: Thanks. nothing is changed and now it looks more beautiful. – Goodarz Mehr Jan 20 '12 at 3:54

## 1 Answer

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, $$\limsup\limits_{n\to\infty}\ R_n=r\cdot\limsup\limits_{n\to\infty}\ u_n=r=\sqrt{c^2+1}.$$

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