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Apparently, the summation $$ \sum_{j = i + 1}^n \frac{1}{j - i + 1} $$ is equal to the summation $$ \sum_{k=1}^{n - i} \frac{1}{k + 1} $$ I don't grasp the intuition behind why.

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    $\begingroup$ Put $j-i=k$ in the first sum. $\endgroup$
    – npisinp
    Apr 9 '14 at 19:36
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    $\begingroup$ Write them out explicitly, and you'll see why. $\endgroup$ Apr 9 '14 at 19:37
  • $\begingroup$ @DavidMitra is right on the money with this (+1). Sometimes I confuse myself with complicated indices, and the only way to be sure is to check that you really end up with exactly the same set of summands when the indices are realized in the sum. $\endgroup$
    – MPW
    Apr 9 '14 at 19:58
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Set $k=j-i$ so when $j=i+1$ then $j-i=1$ so $k=1$. Now when $j=n$ we have $k=j-i=n-i$.

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$$ \sum_{j -i = 1}^n \frac{1}{j - i + 1}\\ \text{Let $j-i=k$, giving }\sum_{k=1}^{n-i} \frac{1}{k + 1}$$

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$\sum_{j = i + 1}^n \frac{1}{j - i + 1}$

$=\frac{1}{2}+...+\frac{1}{n-i+1}$

$=\sum_{k=1}^{n - i} \frac{1}{k + 1}$

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