Proving the limit I have to prove that:$$\lim_\limits{x\to\infty}\bigg(\frac{n}{\frac{1}{x+a_1}+\frac{1}{x+a_2}+\cdots+\frac{1}{x+a_n}}-x\bigg)=\frac{a_1+a_2+\cdots+a_n}{n}$$
The way I started doing this is:
$$=\lim_\limits{x\to\infty}\left(\frac{n}{\frac{(x+a_1)(x+a_2)\cdots(x+a_n)\sum_{i=1}^{n}\big(\frac{1}{x+a_i}\big)}{(x+a_1)(x+a_2)\cdots(x+a_n)}}-x\right)$$
Then I combine $x$ with the rest, but that leads me nowhere. Any tips on how to do this? Taylor expansion cannot be used.
 A: You may write $1/(x+a_1) = 1/x \cdot 1/(1+a_1/x) = 1/x \left(1-a_1/x+O\left(1/x^2\right)\right)$.
Here $g(x)=O(f(x))$ means $g(x)/f(x)$ is bounded as $x\rightarrow \infty$.
So
$$\begin{align}\lim_{x\to \infty}\left(\frac{n}{\sum_{i=1}^n{\frac{1}{x+a_i}}}-x\right)
&=\lim_{x\to \infty}\left(\frac{n}{\frac{1}{x}\sum_{i=1}^n\left({1-\frac{a_i}{x}+O\left(1/x^2\right)}\right)}-x\right)\\
&=\lim_{x\to \infty}\left(\frac{n}{\frac{1}{x}\left(n-\frac{\sum a_i}{x}+O(1/x^2)\right)}-x\right)\\
&=\lim_{x\to \infty}\left(x\frac{1}{\left(1-\frac{\sum a_i}{nx}+O(1/x^2)\right)}-x\right)\\
&=\lim_{x\to \infty}\left(x\left(1+\frac{\sum a_i}{nx}+O(1/x^2)\right)-x\right)\\
&=\lim_{x\to \infty}\left(\frac{\sum a_i}{n}+O(1/x)\right)\\
&=\frac{a_1+\cdots+a_n}{n}
\end{align}$$
A: Hints: Factor out $1/x$ downstairs and use
$$\frac{1}{1+u} = 1-u +\frac{u^2}{1+u}.$$
A: $$\dfrac{n}{\sum_i \dfrac{1}{x+a_i}} - x =  \dfrac{n-\sum_j\dfrac{x}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} =  \dfrac{n-\sum_j\dfrac{x+a_j}{x+a_j}+\sum_j\dfrac{a_j}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} = \dfrac{\sum_j\dfrac{a_j}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} =$$ $$= \sum_j \dfrac{a_j}{\sum_i\dfrac{x+a_j}{x+a_i}}$$
Now 
$$\displaystyle\lim_{x\to\infty} \sum_j \dfrac{a_j}{\sum_i\dfrac{x+a_j}{x+a_i}} = \sum_j \dfrac{a_j}{\sum_i\displaystyle\lim_{x\to\infty}\dfrac{x+a_j}{x+a_i}} = \sum_j \dfrac{a_j}{n} = \dfrac{a_i + \ldots + a_n}{n}$$
