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I am struggling to understand the concept of sigma sum rearrangement. In fact, I don't even know what to call it. That being said, if anybody can recommend sources for me to study this from or let me know what you call this type of problem so that I can study it more, that would be greatly appreciated.

Let $H_{k}$ represent the series of harmonic numbers. Then \begin{align} (1&.) \quad \sum\limits_{1 \leq j \leq k} \frac{1}{j} = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{k} \\ (2&.) \quad \sum\limits_{1 \leq j \leq k} \frac{1}{j} = H_{k} \\ \end{align}

Now let: \begin{align} (3&.) \quad \sum\limits_{1 \leq k \leq n} kH_{k} \\ (4&.) \quad \sum\limits_{1 \leq k \leq n} k \sum\limits_{1 \leq j \leq k} \frac{1}{j} \\ (5&.) \quad \sum\limits_{1 \leq j \leq k \leq n} k \frac{1}{j} \\ (6&.) \quad \sum\limits_{1 \leq j \leq n} \frac{1}{j} \sum\limits_{j \leq k \leq n} k \end{align}

Equation $(4.)$ is just a simple substitution but then I begin to get confused. I am used to the following notation: \begin{align} \sum_{j=1}^{k} \frac{1}{j} \end{align}

Thus, I would represent $(4.)$ as: \begin{align} \sum_{k=1}^{n} k \sum_{j=1}^{k} \frac{1}{j} \end{align}

I am thus confused by equation $(5.)$ as I don't know how to represent it in my usual way.

Now to my real question. Equation $(6.)$ contains two visual changes:
1. $\frac{1}{j}$ and $k$ swap places
2. The indices change from $1 \leq k \leq n$ and $1 \leq j \leq k$ change to $1 \leq j \leq n$ and $j \leq k \leq n$

I'm not sure how these indices are changed? Can anyone explain this to me, please? Is there a general method for this?

Thanks!

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    $\begingroup$ Not an answer, but a strategy. Don't even try to "rearrange $\Sigma$'s". Write out all your expressions with ellipses ($\cdots$) as you did in (1) and see what is actually going on. Then you can change indices and use the distributive law without fear of error. $\endgroup$ – Ethan Bolker May 15 at 16:20
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If you look closely, the equation change from 5 to 6 is the reverse of what's happening in 4 to 5. Let's take a look at what's happening to the indices with an example.

Let $n=3$, and we'll figure out what pairs of $(j,k)$ are valid for the index set in each equation.

In equation $4$, going through in order, we have $(1,1),(1,2),(2,2),(1,3),(2,3),(3,3)$. As long as we have those six pairs, all the sums will be the same.

Going from equation $4$ to equation $5$ is an application of the distributive law. If you wrote it out, it would look like: $$1(\frac{1}{1})+2(\frac{1}{1}+\frac{1}{2})+3(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}) = 1\cdot\frac{1}{1} +2\cdot\frac{1}{1}+2\cdot\frac{1}{2} +3\cdot\frac{1}{1}+3\cdot\frac{1}{2}+3\cdot\frac{1}{3}$$

In equation $6$, the indices are the same again, just in a different order: $(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)$

And everything is distributed by fractions instead of coefficients:

$$1\cdot\frac{1}{1} +2\cdot\frac{1}{1}+2\cdot\frac{1}{2} +3\cdot\frac{1}{1}+3\cdot\frac{1}{2}+3\cdot\frac{1}{3}= \frac{1}{1}(1 +2+3) +\frac{1}{2}(2+3) +\frac{1}{3}(3)$$

You can generalize what's happening for any $n$, but it's the same operations. As a general strategy, when notation gets difficult, expand it by trying small numbers or simple examples.

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  • $\begingroup$ Wow! I instantly understood the problem thanks to only one of your sentences! Perfect explanation, thanks! $\endgroup$ – Imak May 15 at 16:48

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