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At least for $k \neq 0$ and $k \neq 1$, I'm not sure about the rest.

If you look for example at $\sum\limits_{n=1}^{\infty} \frac{1}{2^n}$; if you take any term in that series it will be greater than the sum of any number of subsequent terms, no matter how many.

In this specific case, my visual intuition is that if you look at a line of unit 1 that you split in half, all of the terms after $\frac{1}{2}$ are contained in the second half of the line, and if they added up to 1/2 we would get to the end of the line, namely 1, therefore beyond the limit. Same principle for subsequent terms.

But how would you prove this formally? I also welcome additional intuition and ways to think about why this would be true.

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    $\begingroup$ It's helpful to note when working with geometric series that that $\sum_{k=0}^{n-1} x^k = \frac{1-x^n}{1 - x}$. $\endgroup$ – muaddib Jun 9 '15 at 21:48
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Not a formal proof, but could be easily turned into one:

You can view $\sum_{n=1}^\infty \frac{1}{k^n}$ as the real number whose expression in base (radix) $k$ is $0.11111\ldots$

Clearly $0.1$ is greater than $0.0111\ldots$; $0.01$ is greater than $0.00111\ldots$; etc.

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    $\begingroup$ Very very cool; this is great intuition, thank you. $\endgroup$ – jeremy radcliff Jun 9 '15 at 23:32
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For $k \ge 2$ and $N \ge 0$ fixed, you have $$\sum_{n=N+1}^{+\infty} k^{-n} = k^{-N}\frac{1}{k-1} \le k^{-N}$$ this is exactly what you wanted.

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Note that $$ \sum_{n=m}^M k^n = k^{-m} \sum_{r=0}^{M-m} k^{-r} \\ = \frac{1}{k^m} \frac{1-k^{m-M-1}-1}{1-1/k} \\ = \frac{1}{k^m} \frac{k(1-k^{m-M-1})}{k-1} \\ < \frac{1}{k^m} \frac{k}{k-1} \\ = \frac{1}{k^{m-1}} \frac{1}{k-1} \\ < \frac{1}{k^{m-1}}, $$ using the geometric series formula and that $k>1$. The last line is the term before the first term of the sum. (Note we do need $k>1$, or the series doesn't converge)

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