# Testing if a geometric series converges by taking limit to infinity

If the limit as n approaches infinity of a geometric series is not zero, then that means the series diverges. This makes intuitive sense to me, because it is an infinite series and we keep adding nonzero terms, it will go to infinity.

However, if the limit as n approaches infinity does equal zero, series that is not enough information to tell whether the series converges or diverges.

I would think that if the limit as n approaches infinity is zero, and the series is a continuous function, then the series would converge to some real number. Why is this not the case? The only counter-example I can think of is a series like: a^n = { (-1)^n }, but that is not a continuous function.

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Convergence of infinite series has nothing to do with continuity; sequences are functions defined on the natural numbers, and any such function is continuous since the natural numbers form a discrete set. The fact that we often consider sequences defined by formulas like $a_n = \frac{1}{n^2}$ which happen to make sense for all real values of $n$ is a red herring, and in particular the continuity of the functions determined by such formulas is largely irrelevant. – Paul Siegel Aug 9 '12 at 1:48

## 1 Answer

$\lim_{n \rightarrow \infty} \frac{1}{n} = 0$ but $\sum_{n = 1}^\infty \frac{1}{n}$ does not converge. The associated function $\frac{1}{x}$ is continuous from $[1, \infty)$.

Also a geometric series is the series associated with the sequence $a_n = pr^n$ for some $r$ and $p$. Geometric series converges if $|r| < 1$.

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So if |r| > 1 it automatically diverges? – ordinary Aug 9 '12 at 1:42

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Yes. If $|r| \geq 1$, then $\lim_{n \rightarrow 0} pr^n \neq 0$. – William Aug 9 '12 at 1:45
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