Limit of a complex sequence So I wanted to calculate $$\lim_{n\rightarrow\infty}\frac{n^2}{(4+5i)n^2+(3+i)^n}$$
I thought that I could do it easier if I calculate $\lim_{n\rightarrow\infty}\frac{(3+i)^n}{n^2}$. First I write $\phi=\arctan(\frac{1}{3})$ so that $3+i=\sqrt{10}(\cos\phi+i\cdot\sin\phi)$. Now we have $\lim_{n\rightarrow\infty}\frac{(3+i)^n}{n^2}=\lim_{n\rightarrow\infty}\frac{(\sqrt{10}(\cos\phi+i\cdot\sin\phi))^n}{n^2}=\lim_{n\rightarrow\infty}\frac{10^{n/2}(\cos(n\cdot\phi)+i\cdot\sin(n\cdot\phi))}{n^2}$. Looking now at the limit of the absolute value of the real and imaginary part, we see both go to $\infty$. Knowing that we than have the complex number should go to $\pm\infty\pm i\infty$. Well adding there $4+5i$ doesn't change a lot. If we now look at $\frac{1}{\pm\infty\pm i\infty}$, can we say it equals to $0$? I am still a bit confused with the complex infinity, but in theory it should. Is there maybe a better proof of this limit?

hint: $|3+i|>2$. I hope this helps.
