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Anyone is welcome to come to lab on Tuesday, and it will be zoomed and recorded (so long as I remember; and you can help!).
\[ \sum_{k=1}^\infty\frac{x}{2^k} = x\sum_{k=1}^\infty\frac{1}{2^k} = x \]
No big deal, right?
\[ \sum_{k=1}^\infty x^k \] Well actually, if we let $x=\frac{1}{2}$, this is Zeno's problem: \[ \sum_{k=1}^\infty \left(\frac{1}{2}\right)^k = \sum_{k=1}^\infty \frac{1}{2^k} \]