I need to proof if the following function is bounded and convergent.
$f(n)=\left(\frac{10+in}{n^{2}+2in}\right)^{n}$
Status:
This should be correct. Can anybody confirm this?
I tried it with Bernoulli now:
$\lim_{n\rightarrow\infty}q^{n}=0$ for $\left|q\right|<1,\quad q\in\mathbb{C}$
$ \begin{eqnarray*} \left|\frac{10+in}{n^{2}+2in}\right| & = & \left|\frac{10n^{2}-20in+in^{3}-2i^{2}n^{2}}{n^{4}+4n^{2}}\right|\\ & = & \left|\frac{10n^{2}-20in+in^{3}+2n^{2}}{n^{4}+4n^{2}}\right|\\ & = & \left|\frac{12n^{2}-20in+in^{3}}{n^{4}+4n^{2}}\right|\\ & = & \left|\frac{12n^{2}+in^{3}-20in}{n^{4}+4n^{2}}\right|\\ & = & \left|\frac{12n^{2}}{n^{4}+4n^{2}}+i\frac{n^{3}-20n}{n^{4}+4n^{2}}\right|\\ & = & \sqrt{\left(\frac{12n^{2}}{n^{4}+4n^{2}}\right)^{2}+\left(\frac{n^{3}-20n}{n^{4}+4n^{2}}\right)^{2}}\\ & = & \sqrt{\frac{144n^{4}}{\left(n^{4}+4n^{2}\right)^{2}}+\frac{n^{6}-40n^{4}+400n^{2}}{\left(n^{4}+4n^{2}\right)^{2}}}\\ & = & \sqrt{\frac{n^{6}+104n^{4}+400n^{2}}{n^{8}+8n^{6}+16n^{4}}}\\ & = & \sqrt{\frac{n^{2}(n^{4}+104n^{2}+400)}{n^{4}(n^{4}+8n^{2}+16)}}\\ & = & \sqrt{\frac{(n^{2}+100)(n^{2}+4)}{n^{2}(n^{2}+4)^{2}}}\\ & = & \sqrt{\frac{n^{2}+100}{n^{2}(n^{2}+4)}}\\ & = & \sqrt{\frac{1}{n^{2}}\cdot\frac{1}{n^{2}+4}\cdot(n^{2}+100)}\\ & = & \sqrt{\frac{1}{n^{2}}}\cdot\sqrt{\frac{1}{n^{2}+4}}\cdot\sqrt{n^{2}+100}\\ & = & \frac{\sqrt{\frac{n^{2}+100}{n^{2}+4}}}{n}\\ \Rightarrow\frac{\sqrt{\frac{n^{2}+100}{n^{2}+4}}}{n} & < & 1\\ n_{1} & > & \sqrt{\frac{1}{2}(\sqrt{409}-3)}\\ n_{2} & < & -\sqrt{\frac{1}{2}(\sqrt{409}-3)} \end{eqnarray*} $
The function converges for $n>\sqrt{\frac{1}{2}(\sqrt{409}-3)}$. Therfore it ist also bounded.
$$f(n)=\left(\frac{10+in}{n^{2}+2in}\right)^{n}$$
Gives:
$$f(n)=\left(\frac{\sqrt{n^2+100}e^{(arg(10+in))i}}{\sqrt{n^4+4n^2}e^{(arg(n^2+2in))i}}\right)^{n}=$$
$$f(n)=\left(\frac{\sqrt{n^2+100}}{\sqrt{n^4+4n^2}}e^{((arg(10+in))-(arg(n^2+2in)))i}\right)^{n}=$$
$$f(n)=\sqrt{\frac{{n^2+100}}{{n^4+4n^2}}}^ne^{((arg(10+in))-(arg(n^2+2in)))ni}=$$
$$f(n)=\left(\frac{n^2+100}{n^2(n^2+4)}\right)^{\frac{n}{2}}e^{in\left(-iln\left(\frac{10+in}{|10+in|}\right)+iln\left(\frac{2in+n^2}{|2in+n^2|}\right)\right)}$$