Help simplifying a Sigma sum so it has k=1 on the bottom.

Question

In an analysis lecture this week, my lecturer made the following step as part of a Cauchy Sequence proof but I don't understand how it makes sense:

$\sum_{k=n}^{m-1} (\frac{1}{9})^{k-1} = (\frac{1}{9})^{n-1} \sum_{k=1}^{m-n} (\frac{1}{9})^{k-1} $

Have I miscopied this or is this valid?

If so, could you explain it for me?

Thanks in advance!

Answer 1

What has been done is valid and can be explained as follows:

You are summing over $k$ where $k$ goes from $n$ to $m+1$. Substituting $l:=k-n+1$ (this is called an index shift) makes the sum go from $n-n+1=1$ to $(m-1)-n+1=n-m$.

We thus obtain \begin{align*}\sum_{k=n}^{m-1} \left(\frac{1}{9}\right)^{k-1} & =\sum_{k=n}^{m-1} \left(\frac{1}{9}\right)^{(k-n+1)+n-2} \\ & \overset{l=k-n+1}{=} \sum_{l=1}^{n-m} \left(\frac{1}{9}\right)^{l+n-2}\\ & =\left(\frac{1}{9}\right)^{n-2} \sum_{l=1}^{n-m} \left(\frac{1}{9}\right)^{l}\\ &=\left(\frac{1}{9}\right)^{n-1} \sum_{l=1}^{n-m} \left(\frac{1}{9}\right)^{l-1} \end{align*}

Funnily enough no wikipedia page in english exists for this precise topic, but you can find a comprehensible one in German here https://de.wikipedia.org/wiki/Indexverschiebung

Answer 2

Just index transform $$ i(k) = k - (n-1) = k - n + 1 $$ Then $i(n) = 1$, $i(m-1) = m-n$ and $i = k-n+1 \iff k = n + i - 1$ and we get $$ \sum_{k=n}^{m-1} \left( \frac{1}{9} \right)^{k-1} =\sum_{i=1}^{m-n} \left( \frac{1}{9} \right)^{n+i-2} =\left( \frac{1}{9} \right)^{n-1} \sum_{i=1}^{m-n} \left( \frac{1}{9} \right)^{i-1} $$