I'm tracing through solution for a question I was working on. I don't quite understand how they got to the line which I marked with an arrow (apologies for using an image, I don't have much time left and didn't wanna have to look up how to do sigma notation in LaTeX here)

Worked solution

Text from the image:

The given sum is $$\begin{align*} &7\sum_{k=2}^n \frac1{k-1}-3\sum_{k=2}^n\frac1k-4\sum_{k=2}^n\frac1{k+1}\\ =&7\sum_{k=1}^{n-1} \frac1k-3\sum_{k=2}^n\frac1k-4\sum_{k=3}^{n+1}\frac1k\\ \color{red}{\longrightarrow}\overset{?}=&7\left(1+\frac12\right)-3\left(\frac12+\frac1n\right)-4\left(\frac1n+\frac1{n+1}\right)\\ &=9-\frac7n-\frac4{n+1} \end{align*}$$

So I understand how they got the second line, which is just basically taking the coefficients out and using the change of summation index rule.

What I don't understand is the line marked with the red arrow.
I can see that $1$ comes from $1/1$ when $k=1$ and $1/2$ probably comes from $k=2$ but the summation is from $k=1$ to $k=n-1$

How do we know to use $k=2$ and to stop at $k=2$ for that first term?

  • 2
    $\begingroup$ Just for future reference, sums are not too bad in LaTeX: $\sum_{k=1}^{n} k$ gives you $\displaystyle \sum_{k=1}^{n} \ k$, and $\frac{1}{k}$ gives you $\frac{1}{k}$. $\endgroup$ – Sputnik Jun 20 '11 at 16:19
  • $\begingroup$ Thanks, will use them in the future $\endgroup$ – Arvin Jun 20 '11 at 17:13
  • 1
    $\begingroup$ To add some informality to Luffy's very clear explanation, look at the line before the arrow. Over the part of the index world from $3$ to $n-1$, there is complete cancellation. So one separates out from each sum the part that is not from $3$ to $n-1$, and the sum of these is the answer. $\endgroup$ – André Nicolas Jun 20 '11 at 18:37

$$7\sum_{k=1}^{n-1}\frac{1}{k}=7\left(1+\frac{1}{2}\right)+7\sum_{k=3}^{n-1}\frac{1}{k}$$ $$3\sum_{k=2}^{n}\frac{1}{k}=3\left(\frac{1}{2}+\frac{1}{n}\right)+3\sum_{k=3}^{n-1}\frac{1}{k}$$ $$4\sum_{k=3}^{n+1}\frac{1}{k}=4\left(\frac{1}{n}+\frac{1}{n+1}\right)+4\sum_{k=3}^{n-1}\frac{1}{k}$$ Do you see how to get the marked line now?


You should use that $7=4+3$. If you look carefully, you can see that the line preceding your line with question marks should read something like $$7\sum_{k=3}^{n-1} \frac{1}{k} -3\sum_{k=3}^{n-1} \frac{1}{k} - 4 \sum_{k=3}^{n-1} \frac{1}{k} + 7(1+1/2) - 3(1/2+1/n) -4(1/n+1/(n+1)) = $$ $$ = 0 + 7(1+1/2) - 3(1/2+1/n) -4(1/n+1/(n+1)) .$$


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