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I have a geometric sequence whose n-th term is $11^n$:

$1,11,121,1331,...$

I want to know if the pattern continues forever. I understand that if I sum these numbers up, my common ratio is 11 which is not between -1 and 1 so the sum doesn't converge but I don't see any reason why this pattern would fail to terminate since there are countably infinite multiples of 11.

user130306
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    The sequence can go on forever, even if the sum is infinite. – Minus One-Twelfth May 12 '19 at 22:02
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    what pattern you are refering to? – Julian Mejia May 12 '19 at 22:03
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    If you are looking for symmetry, a quick glance at Pascal's Triangle will show that you get carries for $n=5$ with binomial coefficients $1,5,10,10,5,1$ and that breaks the symmetry. The symmetry here from the symmetric coefficients is eventually hidden by the base ten representation of numbers (though this kind of pattern works for any base). – Mark Bennet May 12 '19 at 22:08
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    @MinusOne-Twelfth thank you, that answered my question – user130306 May 12 '19 at 22:13
  • And, just a side-note: there exist system of numbers, where this infinite sum does have a finite value, and that value is $-\frac1{10}$. Look up $p$-adic numbers. – Berci May 12 '19 at 22:37

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The finite sum $$ 1 + x + x^2 + \cdots + x^n = \frac{x^{n+1} -1 }{x-1} $$ whenever $ x \ne 1$.

So you can substitute $11$ for $x$ for any particular $n$ you like.

You can only sum the series "forever" when $-1 \le x < 1$.

Ethan Bolker
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