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I'm stuck at this. How is RHS rearranged? Is it a change of index?

$$ \sum_{n=1}^{2N} \frac{1}{n} - \sum_{n=1}^{N} \frac{1}{n} = \sum_{n=N+1}^{2N} \frac{1}{n} $$

I'm stuck here:

$$ \sum_{n=1}^{2N} \frac{1}{n} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+ \frac{1}{2N} $$ $$ \sum_{n=1}^{N} \frac{1}{n} =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+ \frac{1}{N} $$ $$ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2N}-(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+ \frac{1}{N})= \frac{1}{2N}-\frac{1}{N}=\frac{-1}{2N} $$ Thanks!

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    $\begingroup$ Your subtraction has to be wrong. You take $N$ terms away from $2N$ terms. How come you are only left with two? $\endgroup$ – almagest May 28 '16 at 16:45
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    $\begingroup$ Your very last equation is nonsense - you're leaving out the terms that are hidden by the $\dots$. $(1+1/2+1/3+1/4+1/5+1/6)-(1+1/2+1/3)=(1/4+1/5+1/6)$. $\endgroup$ – David C. Ullrich May 28 '16 at 16:46
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    $\begingroup$ You made a trivial mistake. "Dots" don't represent the same number of terms in the two places! $\endgroup$ – guestDiego May 28 '16 at 16:47
  • $\begingroup$ Remember that $\frac{1}{2N}$ is not the next term, it is $N$ terms later than $\frac{1}{N}$. The term after $\frac{1}{N}$ is $\frac{1}{N+1}$. $\endgroup$ – Michael Burr May 28 '16 at 17:03
  • $\begingroup$ So this is the right expansion? $ \sum_{n=1}^{2N} \frac{1}{n} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{N-1}+\frac{1}{N}+\frac{1}{N+1}+\frac{1}{N+2}+\dots+\frac{1}{2N} $. And $ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{N-1}+\frac{1}{N}$ is canceled out by $\sum_{n=1}^{N} \frac{1}{n}$? $\endgroup$ – JDoeDoe May 28 '16 at 18:19
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You can also simply derive it from the sigma-notation. You have that $$\sum\limits_{n=1}^{2N}\frac1n = \sum\limits_{n=1}^N\frac1n + \sum\limits_{n=N+1}^{2N}\frac1n, $$ and hence $$\require{cancel}\sum\limits_{n=1}^{2N}\frac1n - \sum\limits_{n=1}^{N}\frac1n = \left(\cancel{\sum\limits_{n=1}^N\frac1n} + \sum\limits_{n=N+1}^{2N}\frac1n\right)- \cancel{\sum\limits_{n=1}^{N}\frac1n} = \sum\limits_{n=N+1}^{2N}\frac1n. $$

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Your reorder is wrong. See here: $$ \sum_{n=1}^{2N} \frac{1}{n} - \sum_{n=1}^{N} \frac{1}{n}= \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2N}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+ \frac{1}{N}\right)= $$ $$= \left(\frac{1}{1}-\frac{1}{1}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\cdots+\left(\frac{1}{N}-\frac{1}{N}\right)+\frac{1}{N+1}+\frac{1}{N+2}+\cdots+\frac{1}{2N}= \sum_{n=N+1}^{2N}\frac{1}{n} $$

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This is what it should be like:

$$\sum_{n=1}^{2N} \frac{1}{n}- \sum_{n=1}^{N} \frac{1}{n}$$ $$= \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+ \frac{1}{2N}\right) -\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots+ \frac{1}{N}\right)$$ $$=\frac{1}{N+1}+\frac{1}{N+2}+\frac{1}{N+3}+\dots+ \frac{1}{2N}$$ $$=\sum_{n=N+1}^{2N} \frac{1}{n}$$

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  • $\begingroup$ Thanks, now I get it. However, is it even possible to rewrite it with change of index in LHS? $\endgroup$ – JDoeDoe May 28 '16 at 17:16

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