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What is the difference between these two series?

$$ \begin{align} 1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\ \\ 1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\ \end{align} $$

I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin series. But other than that, why don't they both yield the same result?

Thanks!

GambitSquared
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    They don't give the same result because one has finitely many terms and the other has infinitely many, assuming that by $O(x^{n+1})$ you mean that the series just continues with $x^{n+1}+x^{n+2}+\cdots$. – Arthur Apr 21 '17 at 21:51
  • If $|r|<1$, then as you take the limit as $n$ approaches infinity, the $r^n\rightarrow 0$ in the second sum and you get the first one. – Michael Burr Apr 21 '17 at 21:52
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    Series $\ne$ finite sum. – Martín-Blas Pérez Pinilla Apr 21 '17 at 21:53
  • @Martín-BlasPérezPinilla What is the difference? – GambitSquared Apr 21 '17 at 21:53
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    It may help also to note that for $|r|<1$ you have $\lim\limits_{n\to\infty} r^n=0$ so $\lim\limits_{n\to\infty} 1+r+r^2+r^3+\dots+r^{n-1}=\lim\limits_{n\to\infty}\frac{r^n-1}{r-1}=\frac{0-1}{r-1}=\frac{1}{1-r}$ – JMoravitz Apr 21 '17 at 21:53
  • Being... different things? In this case the finite sum is a partial sum of the series. – Martín-Blas Pérez Pinilla Apr 21 '17 at 21:55
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    What you're asking is basically a sophisticated variation on "Why is $1+1$ not the same as $1+1+1+1+1$?" The answer is that while the two expressions have a lot of terms in common, one expression has a lot more terms than the other. So it would actually be very remarkable if they were the same. – Arthur Apr 21 '17 at 21:58
  • @Arthur, no both can be finite; $O(x^{n+1})$ just means that the error is within the order of $x^{n+1}$ – GambitSquared Apr 22 '17 at 13:31
  • @ImreVégh In that case, setting "$=\frac{1}{1-x}$" at the end is wrong and very misleading. – Arthur Apr 22 '17 at 14:52

4 Answers4

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Hint:

$$1+x+x^2+\dots+x^n=\frac{x^{n+1}-1}{x-1}$$

$$\frac{x^{n+1}-1}{x-1}=\frac{1-x^{n+1}}{1-x}=\frac1{1-x}-\frac{x^{n+1}}{1-x}=\frac1{1-x}+\mathcal O(x^{n+1})$$

which holds as $x\to0$, since $1-x\to1$.

Move $\mathcal O(x^{n+1})$ to the other side and you'll get

$$1+x+x^2+\dots+x^n+\mathcal O(x^{n+1})=\frac1{1-x}$$

Simply Beautiful Art
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Beware that "$=$" is not a symmetric relation in the presence of asymptotic notation. So while it is true that $$\frac{1}{1-x} = 1+x+x^2+x^3+...+x^n+O(x^{n+1})$$ as $x\to 0$, we do not usually say that $$1+x+x^2+x^3+...+x^n+O(x^{n+1})=\frac{1}{1-x}$$ because the left-hand side of that can be a lot of things other than $\frac{1}{1-x}$.

Some consider the conventional use of $=$ in these contexts to be misguided and misleading, and would rather we write $$\frac{1}{1-x} \in 1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})$$ whose right-hand side is interpreted as a set of functions.

hmakholm left over Monica
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The second (finite) series is a particularization of the first (with a shifted index): $$ 1 + x + x^{2} + x^{3} + \cdots + x^{n} + O(x^{n+1})= \frac{1}{1 - x} $$ means: There exists constant $C > 0$ and $\delta > 0$ such that if $|x| < \delta$, then $$ \left|\frac{1}{1 - x} - (1 + x + x^{2} + x^{3} + \cdots + x^{n})\right| \leq C|x|^{n+1}. \tag{1} $$ By contrast, the second (with $n$ replaced by $n + 1$) reads $$ 1 + r + r^{2} + r^{3} + \cdots + r^{n} = \frac{r^{n+1} - 1}{r - 1} = \frac{1 - r^{n+1}}{1 - r} = \frac{1}{1 - r} - \frac{r^{n+1}}{1 - r}. \tag{2} $$ Equation (2) gives the specific form of the $O(x^{n+1})$, namely $\dfrac{x^{n+1}}{1 - x}$.

Andrew D. Hwang
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The two series actually do produce the same result. You just have to compare them fairly.

To begin with, you've listed $n$ terms in the first series (not including the "big-O" notation) and only $n-1$ terms in the second series. Let's make both series have $n$ terms, as follows. (I'm also putting the "closed expression" side of the equation on the left in order to have the big-O on the right, as is the standard practice.)

\begin{align} \frac{1}{1-x} &= 1+x+x^2+x^3+...+x^n+\mathcal O\left(x^{n+1}\right),\\ \frac{r^{n+1}-1}{r-1} &= 1+r+r^2+r^3+...+r^n. \end{align}

Notice that $\frac{(-y)}{(-y)} = \frac yy,$ so if we reverse the subtraction on the top and bottom of $\frac{r^{n+1}-1}{r-1}$ at the same time, we get the same result: $$ \frac{1 - r^{n+1}}{1 - r} = 1+r+r^2+r^3+...+r^n. $$

Now distribute the numerator of $\frac{1 - r^{n+1}}{1 - r}$ over the denominator: $$ \frac{1}{1 - r} - \frac{r^{n+1}}{1 - r} = 1+r+r^2+r^3+...+r^n. $$

Add $\frac{r^{n+1}}{1 - r}$ to both sides: $$ \frac{1}{1 - r} = 1+r+r^2+r^3+...+r^n + \frac{r^{n+1}}{1 - r}. $$

The first equation occurs in a context where $x$ is very small (close to zero), so that the $\mathcal O\left(x^{n+1}\right)$ term represents just an error correction to the previous terms. If we likewise suppose that $\lvert r\rvert \ll 1,$ so that we are comparing apples to apples (both series dealing with the same kind of numbers), then $\frac{r^{n+1}}{1 - r} = \mathcal O\left(r^{n+1}\right),$ and we can write $$ \frac{1}{1 - r} = 1+r+r^2+r^3+...+r^n + \mathcal O\left(r^{n+1}\right). $$

So this is the same as the first equation, but with the variable $r$ substituted for $x.$

David K
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