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Let $$f(m)=\dfrac{m+1}{\dfrac{m+m+1}{\dfrac{m}{1-x^{m}}-\dfrac{m+1}{1-x^{m+1}}+\dfrac{1}{2}}+\dfrac{m+1+m+2}{\dfrac{m+1}{1-x^{m+1}}-\dfrac{m+2}{1-x^{m+2}}+\dfrac{1}{2}}}$$

Find this limit $$I=\lim_{x\to 1}\dfrac{f(2001)-f(2002)}{f(2002)-f(2003)}$$

My try: since Use (L'Hôpital's rule） we have \begin{align*}&\lim_{x\to 1}\left(\dfrac{n}{1-x^n}-\dfrac{n+1}{1-x^{n+1}}\right)\\ &=\lim_{x\to1}\dfrac{n(1-x^{n+1})-(n+1)(1-x^n)}{(1-x^n)(1-x^{n+1})}\\ &=\lim_{x\to 1}\dfrac{n-nx^{n+1}-n-1+(n+1)x^n}{1-x^n-x^{n+1}+x^{2n+1}}\\ &=\lim_{x\to 1}\dfrac{-n(n+1)x^n+n(n+1)x^{n-1}}{-(n+1)x^n-nx^n+（2n+1)x^{2n}}\\ &=\lim_{x\to 1}\dfrac{-n^2(n+1)x^{n-1}+n(n-1)(n+1)x^{n-2}}{-n(n+1)x^{n-1}-n^2x^{n-1}+2n(2n+1)x^{2n-1}}\\ &=\dfrac{n(n+1)[n-1-n]}{-n^2-n-n^2+4n^2+2n}=-\dfrac{1}{2} \end{align*} This problem is creat a teacher of China, zhejiang university,it is said that teacher want Floored the arrogance of the students.

then I can't,Thank you

math110
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  • I see your problem!This problem needs a trick that will combine both the terms. Or else you end up $1/0$! Can't see the trick but I am sure someone will give a 2 line answer and make us all look stupid! :) – user44197 Jan 02 '14 at 06:32
  • @user44197 I see:). $1-f(m)/f(m+1)$, then $(1-x^m)/（1-x^{m+1})$ – chenbai Jan 02 '14 at 07:18

2 Answers2

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Let $x = 1-z$, we have

$$\begin{align} \frac{m}{1-x^m} &= \frac{m}{1-(1-z)^m}\\ & = \frac{m}{1 - \left(1 - mz + \frac{m(m-1)}{2}z^2 - \frac{m(m-1)(m-2)}{6}z^3 + O(z^4)\right)}\\ & = \frac{1}{z - \frac{m-1}{2}z^2 + \frac{(m-1)(m-2)}{6}z^2 + O(z^3)}\\ & =\frac{1}{z}\left[ 1 - \left( -\frac{m-1}{2}z + \frac{(m-1)(m-2)}{6}z^2\right) + \left(\frac{m-1}{2} z\right)^2 + O(z^3)\right]\\ &= \frac{1}{z} + \frac{m-1}{2} + \frac{m^2-1}{12}z + O(z^2) \end{align} $$ So $$\begin{align} &\frac{m}{1-x^m} - \frac{m+1}{1-x^{m+1}} + \frac12\\ =& \left(\frac{1}{z} + \frac{m-1}{2} + \frac{m^2-1}{12}z\right) - \left(\frac{1}{z} + \frac{m}{2} + \frac{m^2-2m}{12}z\right) + \frac12 + O(z^2)\\ =&-\frac{2m+1}{12} z + O(z^2) \end{align} $$ This gives us $$f(m) = \frac{m+1}{\frac{2m+1}{-\frac{2m+1}{12}z + O(z^2)} + \frac{2m+3}{-\frac{2m+3}{12}z + O(z^2)}} = -\frac{m+1}{24} z + O(z^2) $$ and hence $$\lim_{x\to 1}\frac{f(2001)-f(2002)}{f(2002)-f(2003)} = \lim_{z\to 0}\frac{-\frac{2002}{24} z + \frac{2003}{24} z + O(z^2) }{-\frac{2003}{24} z + \frac{2004}{24} z + O(z^2)} = \lim_{z\to 0}\frac{\frac{z}{24}+O(z^2)}{\frac{z}{24} + O(z^2)} = 1$$

achille hui
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    Actually, this question seems to be created by combining Taylor expansion of $\frac{m}{1-x^m}$ at different $m$ and to make it harder, the prof choose the coefficients to kill the leading terms. The natural approach of this is to use the Taylor expansion. – achille hui Jan 02 '14 at 07:27
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Just for fun, I'm adding this long computation which doesn't involve the Hospital's rule and explains your (...). $$ \lim_{x \to 1}\left(\frac{n}{1-x^n}-\frac{n+1}{1-x^{n+1}}\right) \\ = \lim_{x \to 1} \frac {\frac{n}{1 + x + \cdots + x^{n-1}} - \frac{n+1}{1+x+\cdots+x^n} }{1-x} \\ = \lim_{x \to 1} \frac 1{1+x+\cdots+x^{n-1}} \frac 1{1+x+\cdots +x^n} \frac{n(1 + x + \cdots + x^n) - (n+1)(1+x+\cdots+x^{n-1}) }{1-x} \\ = \frac 1{n(n+1)} \lim_{x \to 1} \frac {nx^n - (1+x+\cdots+x^{n-1})}{1-x} \\ = \frac{-1}{n(n+1)} \lim_{x \to 1} \frac{(x^n - 1) + (x^n - x) + \cdots + (x^n - x^{n-1})}{x-1} \\ = \frac{-1}{n(n+1)} \lim_{x \to 1} \frac{(x^n - 1) + x(x^{n-1} - 1) + \cdots + x^{n-1}(x - 1)}{x-1} \\ = \frac{-1}{n(n+1)} \lim_{x \to 1} \left( (1+x+\cdots+x^{n-1}) + x(1+x+\cdots + x^{n-2}) + \cdots + x^{n-2}(1+x) + x^{n-1}(1) \right) \\ = \frac{-1}{n(n+1)} \lim_{x \to 1} \sum_{i=0}^{n-1} \sum_{j=0}^{n-i} x^{i+j}\\ = \frac{-1}{n(n+1)} \left( n + (n-1) + \cdots + 1 \right) \\ = \frac {-1}2. $$

Patrick Da Silva
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  • Yes,Thank you,But follow is key to solve my problem – math110 Jan 02 '14 at 06:55
  • @china math : I'm essentially showing interest in this question by posting this, and I'm also trying to get others to get started on it by showing some interesting details. I'm still writing this down to try other stuff. – Patrick Da Silva Jan 02 '14 at 07:02