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I have racked my brain to death trying to understand how these two equations are equal:

$$\frac{1}{1-q} = 1 + q + q^2 + q^3 + \cdots$$

as found in http://www.math.dartmouth.edu/archive/m68f03/public_html/partitions.pdf

From what I understand if I substitute $5$ for $q$ the answer:

$$\frac{1}{1-5} = -\frac{1}{4}$$

is much different than:

$$1+5+25+125+\cdots$$

Any help understanding what is going on would be GREATLY appreciated. Thank you all for your time

Michael Hardy
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Parad0x13
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    Your formula is correct only when you plug in values for $q \in [0,1)$. Moreover, you can prove this is correct by considering and rewriting the partial sums of the series on the right-hand side. –  Nov 05 '13 at 03:26
  • So for example when looking to calculate the second equation on page 3 of http://math.berkeley.edu/~mhaiman/math172-spring10/partitions.pdf q cannot be less than 1, unless I've misunderstood something critical – Parad0x13 Nov 05 '13 at 03:26
  • Thank you user710587 for your input, but I'm not sure what 'on the right-hand side' means : ( – Parad0x13 Nov 05 '13 at 03:28
  • The right-hand side of the equation is $1+q+\cdots$ -- See the answer below she's done it for you (or he). –  Nov 05 '13 at 03:30
  • Why did you edit the formula to make it incorrect when it was right to begin with? – anon Nov 05 '13 at 03:34
  • I just added the +q^k and +5^k to show that the equation continues on, I don't think that invalidates the equation, though I may be missing something – Parad0x13 Nov 05 '13 at 03:38
  • You are missing something. The ellipses $\cdots$ are used to indicate the expression goes on infinitely. Ending it with $+q^k$ signifies that it stops at $q^k$ for some number $k$. – anon Nov 05 '13 at 03:42
  • Your right, although either case works for my initial question. The actual reason I appended the 'end' of the equation was because the problem I'm trying to solve stops at some value – Parad0x13 Nov 05 '13 at 03:46
  • The formula for the finite sum $1+q+\cdots+q^k$ is not $\frac{1}{1-q}$ though (see doppz answer for the actual formula), so setting it equal to that is untruthful/inaccurate/misleading. – anon Nov 05 '13 at 03:49
  • I see, thanks for correcting. Will correct the equation! – Parad0x13 Nov 05 '13 at 03:55
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    What you have is not two equations but two expressions. The equation is the statement that the two expressions are equal. – John Bentin Nov 05 '13 at 09:13

4 Answers4

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$S=1+q+q^2+\dots$

For any $k\in \mathbf{N}$,

Let $S_k=1+q+q^2+\dots+q^k$,

Then, $qS_k=q+q^2+q^3+\dots+q^{k+1}$.

So, $(1-q)S_k=S_k-qS_k=1+q+q^2+\dots+q^k-(q+q^2+q^3+\dots+q^{k+1})=1-q^{k+1}$

If $|q|<1$, then $q^k \longrightarrow 0$ as $k \longrightarrow \infty$, so,

$$S=\lim_{k \to \infty}S_k=\lim_{k \to \infty}\frac{1-q^{k+1}}{1-q} =\frac{1}{1-q}$$

Note, however, that the function $$f(x)=\frac{1}{1-x}$$ is valid for all $x \neq 1$. But, $f(x) = 1+x+x^2+\dots$ only for $|x| <1$.

doppz
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    Also, note that the formula for the partial sums (which I would write as $S_k$ to emphasize the dependence on $k$) is correct for any $q$, not just $|q|\lt 1$; the only place you need that condition is for the limit as $k\to\infty$ to make sense. – Steven Stadnicki Nov 05 '13 at 03:33
  • good point, made a couple changes. – doppz Nov 05 '13 at 03:43
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One thing about functions and power series is that they are not always equal when evaluated just anywhere in the complex plane. Generally, given a power series representation of a function, it only even converges at all within a certain radius $|z|<r$ and need not converge outside of it. Within this radius though the power series representation of a holomorphic function should work. (And more generally, we can define a series $\sum c_n(z-a)^n$ around other points $z=a$ whose regions of convergence are various circles or abscissa in the complex plane.)

In particular, the geometric sum formula $\displaystyle\frac{1}{1-z}=1+z+z^2+\cdots$ only holds for $|z|<1$. This is probably the most basic generating function there is. The formula can be proven by analyzing the finite version of the formula for the partial sums and letting the number of terms tend to infinity.

Sometimes it's useful though to treat generating functions as formal power series from $\Bbb C[[x]]$. In this context we do not treat power series as functions at all: we simply think of them as infinite polynomials. Thus even things like $\sum n!z^n$ make sense as formal power series. The ring $\Bbb C[[x]]$ can be defined formally and rigorously using techniques from abstract algebra and topology. (The keywords here are "completion" and "$I$-adic topology.")

anon
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The sum on the RHS of your equation is an infinite geometric series, whose sum is $\frac {1}{1-q}$ only if $|q|$ is less than 1. Your example of $q=5$ is therefore not valid.

Guy Corrigall
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To see why one would think that this equation was true for some $q$, just do the long division of $1-q$ into $1$

DJohnM
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