0
$\begingroup$

Here is a common argument used to prove that the sum of an infinite geometric series is $\frac{a}{1-r}$ (where $a$ is the first term and $r$ is the common ratio): \begin{align} S &= a+ar+ar^2+ar^3+\cdots \\ rS &= ar+ar^2+ar^3+ar^4+\cdots \\ S-rS &= a \\ S(1-r) &= a \\ S &= \frac{a}{1-r} \, . \end{align} I am sceptical about the validity of this argument. It feels like there is something amiss about a proof involving infinite series that makes no mention of the fact that they are typically defined as the limit of their partial sums. The third line involves 'cancelling' all of the terms other than $a$. This makes it seem like an infinite series is actually an infinite string of symbols, rather than a limiting expression. If $a+ar+ar^2+ar^3+\cdots$ is simply a shorthand for $$ \lim_{n \to \infty}\sum_{k=0}^{n}ar^k \, , $$ and $ar+ar^2+ar^3+ar^4+\cdots$ a shorthand for $$ \lim_{n \to \infty}\sum_{k=1}^{n}ar^k \, , $$ then I am struggling to see how the terms actually cancel. Perhaps the third line can be written more formally as \begin{align} S-rS &= (a+ar+ar^2+ar^3+\cdots)-(ar+ar^2+ar^3+ar^4+\cdots) \\ &= \lim_{n \to \infty}\sum_{k=0}^{n}ar^k - \lim_{n \to \infty}\sum_{k=1}^{n}ar^k \\ &= \lim_{n \to \infty}\left(\sum_{k=0}^{n}ar^k - \sum_{k=1}^{n}ar^k\right) \\ &= \lim_{n \to \infty}\left(a +\sum_{k=1}^nar^k - \sum_{k=1}^{n}ar^k\right) \\ &= \lim_{n \to \infty}a \\ &= a \, . \end{align} There is another concern I have. Geometric series only converge when $|r|<1$. However, the argument used above seems to apply regardless of whether $|r|<1$, which would yield nonsensical results such as $$ 1+2+4+8+\cdots = -1 \, . $$ So is the argument rigorous, and if so, why are my fears misplaced?

$\endgroup$
7
  • $\begingroup$ Typo $n+1$ in the third line, invalidating what follows. $\endgroup$
    – user65203
    Commented Dec 18, 2020 at 12:59
  • $\begingroup$ @YvesDaoust I can't see where the typo is, sorry. $\endgroup$
    – Joe
    Commented Dec 18, 2020 at 13:09
  • $\begingroup$ @YvesDaoust Could you please point out where I made a mistake? $\endgroup$
    – Joe
    Commented Dec 20, 2020 at 22:47
  • 1
    $\begingroup$ In the second summation, for $ar + ar^2 + ar^3 + ar^4 + \cdots \,$, the upper bound should be $n+1$. $\endgroup$
    – A.J.
    Commented Dec 21, 2020 at 18:35
  • $\begingroup$ @A.J. Thanks for pointing it out. I'm wondering if this issue can be circumvented in the following way, though. If the partial sums are equal to $$ P=ar+ar^2+ar^3+\cdots+ar^{n+1} $$ then I agree that $$ P=\sum_{k=1}^{n+1} ar^k\, . $$ But then, $rS$ can be given by \begin{align} \lim_{n \to \infty}P&=\lim_{n \to \infty}\sum_{k=1}^{n+1} ar^k \\ &= \lim_{n+1\to\infty}\sum_{k=1}^{n+1}ar^k \\ &= \lim_{N\to\infty}\sum_{k=1}^{N}ar^k \, . \end{align} Since $N$ is a dummy variable, we may relabel it as $n$. Is this reasoning valid? $\endgroup$
    – Joe
    Commented Dec 21, 2020 at 18:47

3 Answers 3

2
$\begingroup$

I think the proofs generally start (or at least they should) with something like "Suppose an infinite geometric series converges to the sum $S \,$", after which all the steps would be justified.

The 'cancelling' you mentioned can be avoided by including an extra step:

\begin{align} S &= a+ar+ar^2+ar^3+\cdots \\ rS &= ar+ar^2+ar^3+ar^4+\cdots \\ rS + a &= a+ar+ar^2+ar^3+\cdots \\ &= S\\ S-rS &= a \\ S(1-r) &= a \\ S &= \frac{a}{1-r} \, . \end{align}

Also, since the proof assumes the existence of $S$, it clearly will not apply if $|r| \ge 1 \,$.

$\endgroup$
2
$\begingroup$

Let

$$S_n=\sum_{k=0}^n ar^n.$$

Then

$$(1-r)S_n=S_n-rS_n=a-ar^{n+1}$$ and

$$S_n=a\frac{1-r^{n+1}}{1-r}.$$

This is absolutely rigorous for any $r\ne1$.

And from this, for $|r|<1$,

$$S=\lim_{n\to\infty}S_n=\frac a{1-r}.$$

Now, for $|r|\ge1$, the infinite sum diverges as does the limit of $r^{n+1}$, so there is no paradox.

$\endgroup$
1
$\begingroup$

Your argument is rigorous if you know that your series converges. Namely you first should know that the series converges and afterwards you may use that method to compute its limit. Prior to know that the series converges the method is wrong since, if the series diverges, in the first step you are subtracting two infinities ($S-rS = \infty - \infty$). Your fears are indeed justified.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .