# 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 ?

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.

• is the limit $0$ ? – prebesk Nov 28 '14 at 21:16
• @prebesk Yes${}{}$ – Adhvaitha Nov 28 '14 at 21:16

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 $i=1,\cdots, k.$

Using the fact that

$$(1+x)^\alpha \sim_0 1+\alpha x$$ we get

$$a_0 \sqrt n + a_1 \sqrt {n + 1} + ... + a_k \sqrt {n + k}=\sqrt n\left(a_0+a_1\sqrt{1+\frac1n}+\cdots+a_k\sqrt{1+\frac kn}\right)\\\sim_\infty \frac12\sqrt n\left(\frac{a_1+\cdots+a_k}{n}\right)=-\frac{a_0}{2\sqrt n}\xrightarrow{n\to\infty}0$$

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 $0$.

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 $0$.