Got confused by the statement as the author does not provide any details regarding this property. So in the book "Concrete Mathematics" the authors state:

An index variable that doesn’t appear in the summand (here j) can simply be eliminated if we multiply what’s left by the size of that variable’s index set (here $n-k$).

Then the authors use this property in evaluating the particular sum:

$ \begin{equation} S_n = \displaystyle\sum\limits_{1 \le j < k \le n}^{}{\frac{1}{k-j}} \end{equation} $ = $ \begin{equation} S_n = \displaystyle\sum\limits_{1 \le j < k+j \le n}^{}{\frac{1}{k}} \end{equation} $ - replacing $k$ by $k+j$

$ \begin{equation} S_n = \displaystyle\sum\limits_{1 \le k \le n}^{} \displaystyle\sum\limits_{1 \le j \le n-k}^{}{\frac{1}{k}} \end{equation} $ -summing first on j

$ \begin{equation} S_n = \displaystyle\sum\limits_{1 \le k \le n}^{}{\frac{n-k}{k}} \end{equation} $ - the sum on $j$ is trivial

I have underscored what makes me so confused. According to the authors' note, this is valid, however I seek for the clear explanation or proof that allow to eliminate the index variable by simply multiplying the summand by the upper bound. Could anyone shed more light on that?

  • $\begingroup$ 2nd line, inner sum: "summing first on $j$". Because the "term" is $\dfrac 1 k$ and does not depend on $j$, summing it on $j$ means to sum $n-k$ times the term $\dfrac 1 k$. Thus: $\Sigma_{1 \le j \le n-k} \dfrac 1 k = (n-k) \dfrac 1 k$. $\endgroup$ – Mauro ALLEGRANZA Jun 19 '20 at 10:38
  • $\begingroup$ Yes, I understood that, but I can't see why it's true and valid. $\endgroup$ – SAT Jun 19 '20 at 10:39
  • $\begingroup$ Yes, everything that I have out there is clear, however I don't understand why the third line is true and works for every summation if the index variable is not present in the summand $\endgroup$ – SAT Jun 19 '20 at 10:44
  • $\begingroup$ I see, but I still can't prove to myself that summing on j when the "term" is not present means multiplying by $n-k$ $\endgroup$ – SAT Jun 19 '20 at 11:14

In the last line the following general rule is used:

$$\sum_{j=1}^m c_k = \underbrace{c_k+\cdots + c_k}_{m \times c_k} = mc_k$$

Nevertheless, there is a mistake in the index $k$. When replacing summation indices it is better not to use the same index as it happened in your book:

So, let $d:= k-j$. Then,

$$1\leq d \leq \color{blue}{n-1} \text{ and } 1 \leq j \leq n-d$$

It follows

$$S_n = \displaystyle\sum\limits_{1 \le j < k \le n}^{}{\frac{1}{k-j}} = \sum_{d=1}^{n-1}\sum_{j=1}^{n-d}\frac 1d = \sum_{d=1}^{n-1}\frac{n-d}{d}$$

  • $\begingroup$ Thank you! Could you refer me where I can find this "general rule"? $\endgroup$ – SAT Jun 19 '20 at 11:23
  • 1
    $\begingroup$ @SAT This is actually a direct consequence of the definition of the sum symbol. If you have $m$ equal summands, then the sum is equal to the $m$-fold of the summand. $\endgroup$ – trancelocation Jun 19 '20 at 11:31

Let $f(\cdot)$ be a function that does not depend on $i.$

We prove that $$\sum_{1\le i \le n} f(\cdot) = nf(\cdot)$$ by induction on $n$.

If $n = 1$, then $$\sum_{1\le i \le 1} f(\cdot) = f(\cdot).$$

Suppose that $$\sum_{1\le i \le n-1} f(\cdot) = (n-1)f(\cdot),$$ then $$\sum_{1\le i \le n} f(\cdot) = \Big(\sum_{1\le i \le n-1} f(\cdot)\Big) + f(\cdot) = (n-1)f(\cdot) + f(\cdot)=nf(\cdot).$$

  • $\begingroup$ Thank you for your answer, it's much more clear now! $\endgroup$ – SAT Jun 19 '20 at 11:22

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