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.
Asked
Active
Viewed 457 times
3 Answers
12
$$\dfrac{n}{\sum_i \dfrac{1}{x+a_i}} - x = \dfrac{n-\sum_j\dfrac{x}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} = \dfrac{n-\sum_j\dfrac{x+a_j}{x+a_j}+\sum_j\dfrac{a_j}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} = \dfrac{\sum_j\dfrac{a_j}{x+a_j}}{\sum_i \dfrac{1}{x+a_i}} =$$ $$= \sum_j \dfrac{a_j}{\sum_i\dfrac{x+a_j}{x+a_i}}$$
Now $$\displaystyle\lim_{x\to\infty} \sum_j \dfrac{a_j}{\sum_i\dfrac{x+a_j}{x+a_i}} = \sum_j \dfrac{a_j}{\sum_i\displaystyle\lim_{x\to\infty}\dfrac{x+a_j}{x+a_i}} = \sum_j \dfrac{a_j}{n} = \dfrac{a_i + \ldots + a_n}{n}$$
Darth Geek
- 12,296
-
how can you get the last step in first line? – chenbai Dec 29 '15 at 00:34
-
@chenbai : Sum of $n$ copies of $1$... – Eric Towers Dec 29 '15 at 03:10
-
@EricTowers $\dfrac{\dfrac{a_1}{x+a_1}+\dfrac{a_2}{x+a_2}}{\dfrac{1}{x+a_1}+\dfrac{1}{x+a_2}}=\dfrac{a_1(x+a_2)+a_2(x+a_1)}{x+a_1+x+a_2}$ – chenbai Dec 29 '15 at 03:55
-
@chenbai $$n - \sum_j\dfrac{x+a_j}{x+a_j} = n - \sum_j 1 = n - n = 0$$ – Darth Geek Dec 29 '15 at 10:02
-
@chenbai $$\dfrac{\sum_j\dfrac{a_j}{x + a_j}}{\sum_i\dfrac{1}{x+a_i}} = \sum_j\dfrac{a_j}{x + a_j}\dfrac{1}{\sum_i\dfrac{1}{x+a_i}} = \sum_j\dfrac{a_j}{\sum_i\dfrac{x+a_i}{x+a_i}}$$ – Darth Geek Dec 29 '15 at 10:08
-
$\dfrac{a_1(x+a_2)+a_2(x+a_1)}{x+a_1+x+a_2}=\dfrac{a_1+a_2}{\dfrac{x+a_1}{x+a_2}+\dfrac{x+a_2}{x+a_1}}$ ? – chenbai Dec 29 '15 at 10:31
-
@chenbai Not quite. What my formula claims is the following (using your example): $$\dfrac{\dfrac{a_1}{x+a_1} + \dfrac{a_2}{x+a_2}}{\dfrac{1}{x+a_1} + \dfrac{1}{x+a_2}} = \dfrac{a_1}{\dfrac{x+a_1}{x+a_1} + \dfrac{x+a_1}{x+a_2}} + \dfrac{a_2}{\dfrac{x+a_2}{x+a_1} + \dfrac{x+a_2}{x+a_2}}$$ – Darth Geek Dec 29 '15 at 10:36
-
OK, I see.thanks! – chenbai Dec 29 '15 at 11:01
8
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
$$\begin{align}\lim_{x\to \infty}\left(\frac{n}{\sum_{i=1}^n{\frac{1}{x+a_i}}}-x\right) &=\lim_{x\to \infty}\left(\frac{n}{\frac{1}{x}\sum_{i=1}^n\left({1-\frac{a_i}{x}+O\left(1/x^2\right)}\right)}-x\right)\\ &=\lim_{x\to \infty}\left(\frac{n}{\frac{1}{x}\left(n-\frac{\sum a_i}{x}+O(1/x^2)\right)}-x\right)\\ &=\lim_{x\to \infty}\left(x\frac{1}{\left(1-\frac{\sum a_i}{nx}+O(1/x^2)\right)}-x\right)\\ &=\lim_{x\to \infty}\left(x\left(1+\frac{\sum a_i}{nx}+O(1/x^2)\right)-x\right)\\ &=\lim_{x\to \infty}\left(\frac{\sum a_i}{n}+O(1/x)\right)\\ &=\frac{a_1+\cdots+a_n}{n} \end{align}$$
KalEl
- 3,297
