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Working: $$\sum_{r=2}^n {1\over r-1}-\sum_{r=2}^n {1\over r+1} $$ $$={1\over1} + {1\over2} + {1\over3}+\cdots+{1\over n-1}$$ $$-\left({1\over3}+\cdots+{1\over n-1}+{1\over n+1}\right)$$ This should then make: $$1+{1\over2}-{1\over n+1}$$ But it is: $$1+{1\over2}-{1\over n}-{1\over n+1}$$ Why is there the $-{1\over n}$ term? Also how do you solve $\sum_{r=1000}^\infty {1\over r-1}-{1\over r+1} $ I haven't the faintest. Thanks |
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We want to find a simple expression for
$$\sum_{r=2}^n \left(\frac{1}{r-1}-\frac{1}{r+1}\right).$$
To get an idea about your lost $-\dfrac{1}{n}$, let us add a few terms together, maybe up to $r=7$, to see what's going on. We get
$$\left(1-\frac{1}{3}\right)+ \left(\frac{1}{2}-\frac{1}{4}\right)+ \left(\frac{1}{3}-\frac{1}{5}\right)+ \left(\frac{1}{4}-\frac{1}{6}\right) +\left(\frac{1}{5}-\frac{1}{7}\right)+ \left(\frac{1}{6}-\frac{1}{8}\right) .$$ The $-\dfrac{1}{7}$ is your missing $-\dfrac{1}{n}$ term. Added: The OP has been expanded to include the reasoning. Somewhat altered, it says that the sum is equal to $$\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-2}+\frac{1}{n-1} \right)-\left(\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1} \right).$$ Everything cancels except the first two terms and the last two. |
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The supposedly correct answer (as written) is wrong, as well. For the first part, I'm going to assume that $n\ge4$ (you can do any other relevant cases you like on your own). Then with some rearranging and reindexing, we see that $$\begin{align}\sum_{r=2}^n\left(\frac1{r-1}-\frac1{r+1}\right) &= \sum_{r=2}^n\frac1{r-1}-\sum_{r=2}^n\frac1{r+1}\\ &= \sum_{r=1}^{n-1}\frac1r-\sum_{r=3}^{n+1}\frac1r\\ &= \left(\sum_{r=1}^2\frac1r+\sum_{r=3}^{n-1}\frac1r\right)-\left(\sum_{r=3}^{n-1}\frac1r+\sum_{r=n}^{n+1}\frac1r\right)\\ &= \left(\sum_{r=1}^2\frac1r+\sum_{r=3}^{n-1}\frac1r\right)-\sum_{r=3}^{n-1}\frac1r-\sum_{r=n}^{n+1}\frac1r\\ &= \sum_{r=1}^2\frac1r+\left(\sum_{r=3}^{n-1}\frac1r-\sum_{r=3}^{n-1}\frac1r\right)-\sum_{r=n}^{n+1}\frac1r\\ &= \sum_{r=1}^2\frac1r-\sum_{r=n}^{n+1}\frac1r\\ &= \left(1+\frac12\right)-\left(\frac1n+\frac1{n+1}\right)\\ &= 1+\frac12-\frac1n-\frac1{n+1}.\end{align}$$ That's the actual correct answer. I will simplify it slightly further to save space: $$\sum_{r=2}^n\left(\frac1{r-1}-\frac1{r+1}\right)=\frac32-\frac1n-\frac1{n+1}\quad (n\geq 4)\tag{#}$$ Another way to see this (using less $\sum$ notation) is $$\begin{align}\sum_{r=2}^n\left(\frac1{r-1}-\frac1{r+1}\right) &= \left(1+\cdots+\frac1{n-1}\right)-\left(\frac13+\cdots+\frac1{n+1}\right)\\ &= \left(1+\frac12\right)+\left(\frac13+\cdots+\frac1{n-1}\right)-\left(\frac13+\cdots+\frac1{n+1}\right)\\ &= \frac32+\left(\frac13+\cdots+\frac1{n-1}\right)-\left(\frac13+\cdots+\frac1{n-1}+\frac1n+\frac1{n+1}\right)\\ &= \frac32+\left(\frac13+\cdots+\frac1{n-1}\right)-\left(\frac13+\cdots+\frac1{n-1}\right)-\frac1n-\frac1{n+1}\\ &= \frac32-\frac1n-\frac1{n+1}\end{align}$$ For the second part, observe first that $$\sum_{r=1000}^\infty\left(\frac1{r-1}-\frac1{r+1}\right) = -\sum_{r=2}^{999}\left(\frac1{r-1}-\frac1{r+1}\right) + \sum_{r=2}^\infty\left(\frac1{r-1}-\frac1{r+1}\right).$$ (Why?) Now, using $(\#)$ twice, we have $$\begin{align}\sum_{r=1000}^\infty\left(\frac1{r-1}-\frac1{r+1}\right) &= -\left(\frac32-\frac1{999}-\frac1{1000}\right) + \sum_{r=2}^\infty\left(\frac1{r-1}-\frac1{r+1}\right)\\ &= -\frac32+\frac1{999}+\frac1{1000}+\lim_{n\to\infty}\left(\sum_{r=2}^n\left(\frac1{r-1}-\frac1{r+1}\right)\right)\\ &= -\frac32+\frac1{999}+\frac1{1000}+\lim_{n\to\infty}\left(\frac32-\frac1n-\frac1{n+1}\right)\\ &= -\frac32+\frac1{999}+\frac1{1000}+\frac32\\ &= \frac1{999}+\frac1{1000}.\end{align}$$ I'll let you justify that the limit is correct for yourself. |
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