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I do not understand how to prove that in these series $1 + r + r^2 + r^3 + ... + r^k$ the term $r^k$ dominates, e.g $r^k > 1 + r + r^2 + r^3 + ... + r^{k-1}$. This one is a part of the following lecture

EDIT: actually it says in the lecture that the sum is at most twice the last term, so $1 + r + r^2 + r^3 + ... + r^{k-1} \leq 2r^k$ if I understood it right.

I will appreciate any help. Thank you.

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    $1+r+r^2+\cdots+r^{k-1}=\frac{r^k-1}{r-1}$. Geometric series. https://en.wikipedia.org/wiki/Geometric_series – 2017-01-13
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    Does not seem to hold in general, try $r=\frac{1}{2}$. You probably missed some details. – 2017-01-13
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    $r^k\ge1+r+r^2+r^3+\dotsb+r^{k-1}$ holds if $r>2$. – 2017-01-13
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    The rule in the lecture, sum is bounded by twice the last term, is the same. That is, adding $r^k$ on both sides you get $1+r+r^2+…+r^k\le 2·r^k$ for $r\ge 2$. – 2017-01-13

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As long as $r\ge2$, we have: \begin{align} r^k&>r^k-1\\ &\ge\frac{r^k-1}{r-1}\quad\text{(since $r\ge2$)}\\ &=1+r+r^2+\dotsb+r^{k-1} \end{align}

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If $r=1/2$ or $1$, the statement is not true. If $r$ is an integer greater than $1$, then you're comparing two numbers written in base $r$: $(111\cdots 1)_r$ and $(200\cdots 0)_r$, where the second number has one more digit than the first. If $r$ is real, we need some restrictions on $r$ to make the statement true.