The limit of a sum of the form $a_0 \sqrt n + a_1 \sqrt {n + 1} +\cdots + a_k \sqrt {n + k}$ If $ a_0 ,a_1 ,\ldots,a_k  $ are real numbers such that 
$$ a_0  + a_1  + \cdots + a_k  = 0$$ 
Find $$
 \lim_{n \to \infty } (a_0 \sqrt n  + a_1 \sqrt {n + 1}  +\cdots + a_k \sqrt {n + k} )$$ 
I just can't see any solution, can anyone help ?
 A: We have $a_k = -a_0-a_1-\cdots-a_{k-1}$. Hence,
$$\sum_{l=0}^k a_l \sqrt{n+l} = \sum_{l=0}^{k-1} a_l\left(\sqrt{n+l} - \sqrt{n+k}\right) = \sum_{l=0}^{k-1}a_l \dfrac{l-k}{\sqrt{n-l}+\sqrt{n-k}}$$
Now you should be able to conclude the answer.
A: Since $a_0=-a_1-\cdots -a_k$ we have that
$$\lim_{n \to \infty } (a_0 \sqrt n  + a_1 \sqrt {n + 1}  + ... + a_k \sqrt {n + k} )=\lim_{n \to \infty } (a_1(\sqrt {n + 1}-\sqrt{n})+\cdots + a_k(\sqrt {n + k}-\sqrt{n}))\\=\lim_{n \to \infty } \left(\frac{a_1}{\sqrt {n + 1}+\sqrt{n}}+\frac{2a_2}{\sqrt {n + 1}+\sqrt{n}}+\cdots + \frac{ka_k}{\sqrt {n + k}+\sqrt{n}}\right)=0,$$
where we have used that $$\sqrt{n+i}-\sqrt{n}=\frac{(\sqrt{n+i}-\sqrt{n})(\sqrt{n+i}+\sqrt{n})}{\sqrt{n+i}+\sqrt{n}}=\frac{i}{\sqrt{n+i}+\sqrt{n}},$$ for $i=1,\cdots, k.$
A: Using the fact that
$$(1+x)^\alpha \sim_0 1+\alpha x$$
we get
$$a_0 \sqrt n  + a_1 \sqrt {n + 1}  + ... + a_k \sqrt {n + k}=\sqrt n\left(a_0+a_1\sqrt{1+\frac1n}+\cdots+a_k\sqrt{1+\frac kn}\right)\\\sim_\infty \frac12\sqrt n\left(\frac{a_1+\cdots+a_k}{n}\right)=-\frac{a_0}{2\sqrt n}\xrightarrow{n\to\infty}0$$
A: Substitute $n=\frac{1}{p}$. Then:
\begin{equation}
\lim_{p \to 0} (\frac{a_0}{\sqrt{p}}+\frac{a_1\sqrt{1+p}}{\sqrt{p}}+\ldots+\frac{a_k\sqrt{1+kp}}{\sqrt{p}})
\end{equation}
\begin{equation}
\lim_{p \to 0} (\frac{a_0+a_1\sqrt{1+p}+\ldots+a_k\sqrt{1+kp}}{\sqrt{p}})
\end{equation}
This takes $\frac{0}{0}$ form. Thus, apply L'Hospital rule.
Differentiating both numerator and denominator w.r.t $p$
\begin{equation}
\lim_{p \to 0} \sqrt{p}(\frac{a_1}{\sqrt{1+p}}+\ldots+\frac{ka_k}{\sqrt{1+kp}})
\end{equation}
which equals $0$.
A: Subtract the constrain equation times $\sqrt n$ from the target expression, and you'll see such terms appear:
$$a_i(\sqrt{n+i}-\sqrt n).$$
These can be written as
$$a_i\frac{i}{\sqrt{n+i}+\sqrt n},$$
and obviously then to $0$.
