3
$\begingroup$

Why does the power series of $x + x^2 + x^3\ldots$ not equal to $x/(1-x)$ when $x$ is larger than $1$?

It is never specified what range of values $x$ takes on, so the algebra should work out in all cases. Why does it not work out when $x \gt 1$?

$\endgroup$
  • 4
    $\begingroup$ The series does not converge (in any traditional sense) if $x>1$. $\endgroup$ – lulu Apr 19 '16 at 1:30
  • $\begingroup$ Does it make sense to add up those terms if $x>1$? For instance you might have $2+4+8+16+\cdots$ - does this converge? $\endgroup$ – Cameron Williams Apr 19 '16 at 1:30
  • $\begingroup$ It doesn't, but nonetheless the algebra says that they are equal to each other. $\endgroup$ – Goldname Apr 19 '16 at 1:30
  • $\begingroup$ What algebra are you talking about? $\endgroup$ – Rick Sanchez Apr 19 '16 at 1:31
  • $\begingroup$ The algebra does if you are not careful, but the thing is that you are manipulating infinities by the very nature of it. You cannot make any real sense of subtracting infinities. $\endgroup$ – Cameron Williams Apr 19 '16 at 1:31
5
$\begingroup$

If the manipulations you're talking about are on the order of the following—

$$ S = x + x^2 + x^3 + \cdots $$

$$ xS = x^2 + x^3 + \cdots $$

$$ S-xS = x $$

$$ S = \frac{x}{1-x} $$

—then your issue is in assuming that you can subtract an infinite quantity like $xS$ from another infinite quantity like $S$, and end up with something meaningful (in an ordinary sense). The straightforward interpretation is that subtracting infinities like that is prone to yielding nonsense like $0 = \infty$, so we disallow it.

To be sure, it is possible to stretch the rules so as to make sense of what you're proposing, but the rules are not stretched willy-nilly, but according to new rules. At any rate, I hope that explains why the resulting expression for $S$ does not work in a conventional sense outside the region of convergence $|x| < 1$.

$\endgroup$
2
$\begingroup$

We know that for $|x|<1$, the function $f(x)=\frac{x}{1-x}$ can be written as

$$f(x)=\sum_{n=1}^\infty x^n \tag 1$$

Note that this series is a representation of the function $f(x)$ that is valid for values of $x\in (-1,1)$. However, while the function $f(x)$ is defined for all $x\ne 1$, the series in $(1)$ does not even converge for $|x|\ge 1$.

We can develop, in fact, a series representation for $f(x)$ that is valid for $|x|>1$, in terms of inverse powers of $x$. To do this, we write

$$\begin{align} f(x)&=\frac{x}{1-x}\\\\ &=\frac{1}{1/x-1}\\\\ &=-\frac{1}{1-1/x}\\\\ &=-\sum_{n=0}^\infty \left(\frac{1}{x}\right)^{n} \tag 2 \end{align}$$

where $(2)$ is valid for $|x|>1$. Therefore, we can represent $f(x)$ as

$$f(x)=\begin{cases} \sum_{n=1}^\infty x^n&,|x|<1\\\\ -\sum_{n=0}^\infty \left(\frac{1}{x}\right)^{n} &,|x|>1\\\\ -1/2&,x=-1 \end{cases}$$

$\endgroup$
2
$\begingroup$

A summation will converge only if its individual terms decay to 0 as $n\to\infty$.

$\endgroup$
  • $\begingroup$ I was more hoping of receiving an answer of why the mathematical manipulations didn't work out. $\endgroup$ – Goldname Apr 19 '16 at 1:46
2
$\begingroup$

One can start from calculating a partial summation. $$S_m=\sum_{n=1}^{m} x^n$$ $$S_m (1-x) = \sum_{n=1}^{m} x^n - \sum_{n=1}^{m} x^{n+1} = x-x^{n+1}$$ Therefore, if $x\neq 1 $ $$S_m=\frac{x-x^{n+1}}{1-x}$$ $$\sum_{n=1}^{\infty} x^n = \lim_{m->\infty} S_m$$ You can see that the limit converges only if $|x|<1$

$\endgroup$
1
$\begingroup$

If you view $\sum_{k=1}^\infty x^k$ as a formal power series over $\mathbb R$, then the algebraic manipulations, including the last step of dividing by $1-x$, are all valid within the ring of formal power series. When you try to interpret these power series as functions of $x$, however, you have to be careful about where and how they converge for those manipulations to lead to a valid result.

The above series only converges for $|x|<1$, i.e., when the power series is interpreted as a function, the function is defined only on this interval. Remember that for functions to be equal, not only do their values have to agree, but they also have to have the same domain and range, so it can only be equal to $\frac x{1-x}$ in this limited region.

$\endgroup$
1
$\begingroup$

As a formal power series, we have $x + x^2 + x^3+\cdots=x/(1-x)$.

As a numerical power series, this fails for $|x| > 1$.

The main reason is an important theorem about numerical power series:

If a power series converges for $x=x_0$, then it converges for all $x$ such that $|x| < |x_0|$.

Since $x + x^2 + x^3+\cdots$ clearly does not converge for $x=1$, it cannot converge for any $x>1$.

$\endgroup$
  • $\begingroup$ I find it just a little strange that you invoke the theorem on the domain of convergence of a power series to deduce the $\sum_n x^n$ diverges for $x > 1$. For $x > 1$, $\sum_{n=1}^N x^n \geq \sum_{n=1}^N 1^n$, so the divergence of the first series seems just as clear as the divergence of the second one. $\endgroup$ – Pete L. Clark Dec 3 '16 at 7:38
1
$\begingroup$

We can use the definition of the radius of convergence to show that we only have convergence on the disk $|x| < 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.