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I am reading a book "A History of Mathematics by Boyer" In the chapter about Euler it states that

"Although on occasion he warned against the risk in working with divergent series, he himself used the binomial series"

$1/(1-x) = 1 + x + x^2 + x^3 ...$ for values of $x\geq 1$.

But what about dividing with zero? Isn't it zero when $x =1$ ?

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    $\begingroup$ $1/(1-x)$ series expansion is not what you have written. $\endgroup$ – lebesgue Jun 3 at 6:35
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    $\begingroup$ What you wrote is wrong. It is $$\frac1{1-x}=1+x+x^2+\ldots+x^n+\ldots$$ only for $\;|x|<1\;$ ... $\endgroup$ – DonAntonio Jun 3 at 6:41
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    $\begingroup$ This is not a binomial series. Did you copy the whole text ? $\endgroup$ – Yves Daoust Jun 3 at 7:26
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    $\begingroup$ I am voting to close this question since it is absolutely not clear what your problem is. I showed you below that your original equality is wrong and I gave you a source where the formula for the geometric series is proven. What you are missing in the text is something what $you$ should tell us. $\endgroup$ – Jan Jun 3 at 7:53
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    $\begingroup$ @Asim I'm not sure how I could not be sure. That is what is written. Furthermore, as pointed out numerous times here, it is incorrect. $\endgroup$ – superckl Jun 3 at 8:51
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This can't be true. Note that

$$\frac{1}{1 - 2} = -1,$$

but

$$\sum_{k = 0}^\infty \frac{1}{2^k} = 2.$$

Especially, $1/(1 - x)$ is not defined when $x = 1$.

The correct expansion is

$$\frac{1}{1 - x} = \sum_{k = 0}^\infty x^k, \qquad \lvert x \rvert < 1.$$

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  • $\begingroup$ What exactly arises your confusion? $\endgroup$ – Jan Jun 3 at 7:37
  • $\begingroup$ I updated the question it was my mistake i wrote it wrong. but the confusion is still valid. How come 1/(1-x) becomes 1 + x + x^2... etc for values of x equal to 1 ? $\endgroup$ – Asim Jun 3 at 7:43
  • $\begingroup$ As I have written above: The expression $1/(1 - x)$ does not (!) equal $1 + x + x^2 + \dots$ for $x = 1$. If you want to know why you have equality for $\lvert x \rvert < 1$, just look for geometric series, e.g. in this answer: math.stackexchange.com/questions/29023/… $\endgroup$ – Jan Jun 3 at 7:46
  • $\begingroup$ This is what i am trying to find out. Why does the book says it so. Did u read the screenshot of book i posted. Thanks for the help. $\endgroup$ – Asim Jun 3 at 7:51
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From the formula $$ \sum_{n=0}^∞ x^n = \frac{1}{1-x} $$ valid for $|x|< 1$, you can replace $x$ by $\frac{1}{x}$ and you get that for any $|x|>1$ it holds $$ \sum_{n=0}^∞ \frac{1}{x^n} = \frac{1}{1-\frac{1}{x}} = \frac{x}{x-1} $$ In both of these series however, $|x|≠ 1$.

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