# Find an expression, in terms of n, for $\sum_{r=2}^n {1\over r-1}-{1\over r+1}$

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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I'd edit but I have no idea of what you are asking. – Ron Gordon Jan 7 '13 at 22:02
Do you mean $\sum_{r=2}^n \frac {1}{r-1} - \frac {1}{r+1}$? – Calvin Lin Jan 7 '13 at 22:04
yes! so sorry trying to format properly now! – maxmitch Jan 7 '13 at 22:05
Check to see if my edit still asks what you are wanting to ask. – Cameron Buie Jan 7 '13 at 22:25
You have an error in your work -- $\sum_{r=2}^n \frac{1}{r+1} = \frac{1}{3} + \ldots + \frac{1}{n-1} + \frac{1}{n} + \frac{1}{n+1}$, so the "missing" $\frac{1}{n}$ is explicit... – gt6989b Jan 7 '13 at 23:27

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 $$\left(1-\frac{1}{3}\right)+ \left(\frac{1}{2}-\frac{1}{4}\right)+ \left(\frac{1}{3}-\frac{1}{5}\right)+ \left(\frac{1}{4}-\frac{1}{6}\right) +\left(\frac{1}{5}-\frac{1}{7}\right)+ \left(\frac{1}{6}-\frac{1}{8}\right) .$$
There is a lot of cancellation. Everything disappears except for the $1$, the $\dfrac{1}{2}$, the $-\dfrac{1}{7}$, and the $-\dfrac{1}{8}$.

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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This doesn't really help as I am looking for where the -${1\over7}$ /-${1\over n}$ belongs in comparison to my "working". For example how can I tell that -${1\over n}$ is -${1\over 7}$ – maxmitch Jan 7 '13 at 22:55
I have added something to deal with your added detail, which is correct. The only problem with it is that in the expansion at the end, you did not write in enough terms to see exactly what cancels. I have written the same thing as you did, but with a couple of extra terms, and then everything is (I hope) clear. – André Nicolas Jan 7 '13 at 23:09

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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Thank you! How come $$\sum_{r=1000}^\infty\left(\frac1{r-1}-\frac1{r+1}\right) = -\sum_{r=2}^{999}\left(\frac1{r-1}-\frac1{r+1}\right) + \sum_{r=2}^\infty\left(\frac1{r-1}-\frac1{r+1}\right).$$ ? – maxmitch Jan 7 '13 at 23:15
@maxmitch: Given integers $m, k$ with $m\le k$, and any sequence $a_m, a_{m+1}, a_{m+2},...$ of numbers such that $\sum_{r=m}^\infty a_r$ exists, we have that $$\sum_{r=m}^\infty a_r= \sum_{r=m}^k a_r+ \sum_{r=k+1}^\infty a_r.$$ You should be able to use this general statement to answer your own question. – Cameron Buie Jan 7 '13 at 23:43