Stuck with a sum problem (change of lower bound)

Question

This is from a solution to a problem (3b here). I can't understand why the highlighted parts are equal. I've figured out that $\sum_{i=a}^b i$ is the same as $\sum_{i=1}^{b-a+1} (i+b-a)$. The bounds on the second sum agree with that.

$b$ and $1$ (in the numerator) are constants and may be pulled out of the sum (like $\frac1{b-a+1}$ was), but why have they disappeared altogether? And why has $k$ changed sign?

I think the highlighted part should become $\sum_{ k= 1}^{b-a} (b-k+1+a)$.

What am I missing here?

Thank you in advance.

The highlighted parts are not equal. On the second line, you are missing the factor of $\frac{1}{b-a+1}$ in your circle. — Xander Henderson, Jan 28 '18 at 01:23
@XanderHenderson Well, to me it seems he made sure the circle evades the factor... — The Phenotype, Jan 28 '18 at 01:27
@ThePhenotype Oi... indeed, maybe in the top line it does... it is a bit hard for me to follow. :\ — Xander Henderson, Jan 28 '18 at 01:28

score 3 · Answer 1 · answered Jan 28 '18 at 01:25

3

Let $l=b-k+1$, hence $k=b-l+1$.

when $k$ increases, $l$ decreases.

If $k=b$, then $l=1$.

If $k=a+1$, then $l=b-a$.

\begin{align} \sum_{k={a+1}}^b (b-k+1) &= \sum_{l=1}^{b-a} l \end{align}

answered Jan 28 '18 at 01:25

Siong Thye Goh

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score 1 · Accepted Answer · answered Jan 28 '18 at 01:39

There is property of change of order of the sum $(1)$ $$\sum^{b}_{k=a}f(k)=f(a)+f(a+1)+\cdots+f(b)=$$ $$= f(b)+f(b-1)+\cdots+f(a)=\sum_{k=a}^bf(b+a-k) $$

And other that helps is $(2)$ $$(2)\sum^{b}_{k=a}f(k)=\sum^{b+p}_{k=a+p}f(k-p) $$

$$(1)\sum^{b}_{k=a}f(k)=\sum_{k=a}^bf(b+a-k) .$$

First apply the property $(2)$, subtracting $a$ from the limits

$$\sum^{b}_{k=a+1}(b-k+1)=\sum^{b-a}_{k=1}(b-(a+k)+1)= \sum^{b-a}_{k=1}(b-a+1-k)$$

Then we use the property $(1)$ to change the order $$=\sum^{b-a}_{k=1}k. $$

Stuck with a sum problem (change of lower bound)

2 Answers2