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enter image description here

Each of the purple squares has 1/4 of the area of the next larger square (1/2×1/2 = 1/4, 1/4×1/4 = 1/16, etc.). The sum of the areas of the purple squares is one third of the area of the large square.

https://en.wikipedia.org/wiki/Geometric_series#/media/File:GeometricSquares.svg

I'm wondering whether the geometric series shown is Cauchy. It eventually reaches a limit that is on the upper right hand corner. Converges to the black dot on the upper right hand corner.

Similarly, if the sequence shown below continues and approaches the origin, the center of the picture, I think the picture below is "pell series"?, does the "pell series" converge because it is Cauchy and reaches the center of the graph eventuall?

enter image description here

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  • $\begingroup$ Yes, because it's converges.If $-1<q<1$ then the sequence is a Cauchy sequence. $\endgroup$
    – Michael Rozenberg
    Commented Jun 24, 2017 at 12:26
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    $\begingroup$ Any convergent series is Cauchy ... $\endgroup$
    – Noah Schweber
    Commented Jun 24, 2017 at 12:31

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hint

For $n,p>0$, Let $$S_n=1/4+1/4^2+...1/4^n . $$

$$S_{n+p}-S_n=$$ $$1/4^{n+1}\Bigl(1+1/4+...1/4^{p-1}\Bigr) $$

$$=1/4^{n+1}\frac {1-1/4^p}{1-1/4} $$

$$\le \frac {1}{3.4^n}. $$

thus

$$\forall \epsilon>0 \;\; \exists N\in\mathbb N \;\;: \forall n>N \;\;\forall p>0$$ $$|S_{n+p}-S_n|\le \frac {1}{3.4^n}<\epsilon $$

because $\lim_{+\infty}\frac {1}{3.4^n}=0$. $(S_n)$ is Cauchy...

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  • $\begingroup$ Where does the $3.4^n$ in the denominator come from? Or is that meant to be $(3/4)^n$? $\endgroup$
    – Michael A
    Commented Aug 24, 2022 at 21:34

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