sum of infinite series with a special series

Question

I can find an answer by finding $a_0$ and $a_1$ when $m = 1$ and using them to calculate the sum of infinite series. But I can't generalize for any m. $$\text{for constant}\, a_k,\,(k = 0,\, 1,\, 2,\, ...\, m)$$ $$\sum^m_{k = 0}a_k = 0,\,\sum^m_{k = 1} ka_k=-6$$ $$\text{what is the value of}\, \lim_{n\to\infty}\sum_{k = 0}^m a_k\sqrt{n^2 + kn}\,\text?$$

Answer 1

Since $\sum_{k=0}^ma_k=0$, we have $\sum_{k=0}^mna_k=0$ for any $n$, hence

$$\sum_{k=0}^ma_k\sqrt{n^2+kn}=\sum_{k=0}^ma_k\left(\sqrt{n^2+kn}-n\right)=\sum_{k=0}^ma_k\left(kn\over\sqrt{n^2+kn}+n \right)\to\sum_{k=0}^m{a_kk\over2}={-6\over2}$$

Answer 2

You have: $$\sqrt{n^2+kn}=n \sqrt{1+\frac{k}{n}}=n \left(1+\frac{k}{2 n}+o_{n \to \infty} \left(\frac{1}{n} \right) \right)=n+\frac{k}{2}+o_{n \to \infty}(1)$$ so: $$\sum_{k=0}^m a_k \sqrt{n^2+kn} = n \sum_{k=0}^m a_k+\frac{1}{2} \sum_{k=0}^m k a_k+ o_{n \to \infty}(1)$$ and the limit is thus $-3$

The important point is that the sum is finite (and independent of $n$) you can do all the manipulations with $o_{n \to \infty}$ without any problem.