Let $a(k)=(2k+3)(k-1)$ for every $k\in\mathbb N$. Find: $$\lim_{n \to \infty} \frac{\sqrt{a(4^0n)}+\sqrt{a(4^1n)}+\cdots+\sqrt{a(4^{10}n)}}{\sqrt{a(2^0n)}+\sqrt{a(2^1n)}+\cdots+\sqrt{a(2^{10}n)}}$$
I tried to convert it into limit as a sum form but it isn't fruitful. Need hints.
Here is a bit longer version of @Did's comment above. Note that $$a(k)=2k^2+2k-3=2k^2+o(k^2).$$ (If you haven't seen little o notation, it essentially means that all I care about in the asymptotics is the leading term on $k^2$). Thus we have (asymptotically) \begin{align*} \frac{\sqrt{a(4^0n)}+\dots+\sqrt{a(4^{10}n)}}{\sqrt{a(2^0n)}+\dots+\sqrt{a(2^{10}n})}&\sim\frac{\sqrt{2(4^0n)^2}+\dots+\sqrt{2(4^{10}n)^2}}{\sqrt{2(2^0n)^2}+\dots+\sqrt{2(2^{10}n)^2}}\\ &=\frac{\sqrt{2}n\left(4^{0}+\dots+4^{10}\right)}{\sqrt{2}n\left(2^{0}+\dots+2^{10}\right)}\\ &=??? \end{align*}
Here is a simpler example to possibly help you see what's going on here. Let's try to compute $$\lim_{n\to\infty}\frac{a(4n)}{a(2n)}.$$ Just by using the definitions of the sequence we have $$\lim_{n\to\infty}\frac{a(4n)}{a(2n)}=\lim_{n\to\infty}\frac{2(4n)^2+2(4n)-3}{2(2n)^2+2(2n)-3}=\lim_{n\to\infty}\frac{32n^2+8n-3}{8n^2+4n-3}.$$ What do we do from here? Let's multiply both numerator and denominator by $\frac{1}{n^2}$. We have $$\lim_{n\to\infty}\frac{a(4n)}{a(2n)}=\lim_{n\to\infty}\frac{32+\frac{8}{n}-\frac{3}{n^2}}{8+\frac{4}{n}-\frac{3}{n^2}}.$$ Can you see now what this limit should be, and why @Did and I only cared about the leading term in the original example? They work exactly the same way.