A supposedly fast method to find the sum of a geometric series is the following one.

Let $$S = \sum_{n = 0}^{+\infty} q^n$$ then $$S = 1 + q\left[\sum_{n = 0}^{+\infty} q^n\right] = 1 + qS.$$ Hence $$(1 - q)S = 1 \implies S = \frac1{1 - q}$$

This result, apparently correct, is actually wrong, since we have never required $|q| < 1$, and so it would seem that the geometric series converges for every value of $q$, which is not true.

I cannot find where the error lies in the derivation of the above result. Any hints?

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    $\begingroup$ The error is giving the "sum" a name and manipulating that name as if it were a number. $\endgroup$ – André Nicolas Jul 7 '15 at 5:30
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    $\begingroup$ You started wrong. The expression S is only valid for |q|<1, otherwise it is inifinite (neg or pos), so your algebraic manipulation is all bogus after that. No banana. $\endgroup$ – user247608 Jul 7 '15 at 5:31
  • $\begingroup$ The error is in the second step I guess, while defining associativity of real numbers, doing $q(\sum_i a_i) = \sum_i (qa_i)$ is allowed only when the sum $\sum_i$ is finite. There is no guarantee in general that $\sum_i a_i$ was a finite real number in the first place. $\endgroup$ – r9m Jul 7 '15 at 5:31
  • $\begingroup$ @AndréNicolas but how is this different from this calculation? There the same thing is done. $\endgroup$ – rubik Jul 7 '15 at 5:32
  • $\begingroup$ @rubik: There is no issue about the existence of finite sums, which is what is being worked with in the linked article. $\endgroup$ – André Nicolas Jul 7 '15 at 5:35

By multiplying $q$ by $S$ you implicitly assume that the series converges, because otherwise you would be multiplying by infinity, which doesn't work.

  • $\begingroup$ In response to your comment above, as I lack the reputation to reply, that is taking the sum of a finite number of terms. Taking the limit as n approaches infinity gives you a result of either infinity (if it diverges) or a value (if it converges). $\endgroup$ – anon249759374 Jul 7 '15 at 5:37

If $S = \infty$ you perform a "bad" operation:

$S = 1 + q S$ reads $\infty = 1 + q \infty$ you cannot compute $\infty - q \infty = (1-q)\infty$ since $\infty - \infty$ makes no sense


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