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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 $x =1$ ?

enter image description here

Asim
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    $1/(1-x)$ series expansion is not what you have written. – Dinesh Jun 03 '20 at 06:35
  • The equation is not correct. When $x=2$ the series has sum $2$ whereas LHS is $-1$. – Kavi Rama Murthy Jun 03 '20 at 06:35
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    What you wrote is wrong. It is $$\frac1{1-x}=1+x+x^2+\ldots+x^n+\ldots$$ only for $;|x|<1;$ ... – DonAntonio Jun 03 '20 at 06:41
  • If you expand $\dfrac 1 {1 - 1/x}$ you do get what OP wrote. Multiplying top and bottom of the LHS you get $\dfrac x {x - 1} = 1 + 1/x + 1/x^2 + 1/x^3 + \cdots$ which may help to explain OP's confusion -- but is valid for $\left|{\dfrac 1 x}\right| < 1$, that is, $|x| > 1$. (Please check that I've got this right.) – Prime Mover Jun 03 '20 at 06:46
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    This is not a binomial series. Did you copy the whole text ? –  Jun 03 '20 at 07:26
  • What it says in the book is "he himself used the binomial series $1/(1-x) = 1+x+x^2+x^3+\cdots$ for values of $x \ge 1$. In fact, by combining the two series $x/(1-x)=x+x^2+x^3+\cdots$ and $x/(x-1)=1+1/x+1/x^2+\cdots$ Euler concluded that $\cdots +1/x^2+1/x+1+x+x^2+x^3+\cdots=0$." This is the 2nd edition by Boyer & Merzbach, 1991. – Prime Mover Jun 03 '20 at 07:31
  • Prime Mover would u explain further. What am i missing in the text. – Asim Jun 03 '20 at 07:37
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    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. – Jan Jun 03 '20 at 07:53
  • What i have asked is not MY equation. It is equation written in the book which I am trying to understand because my previous knowledge of math says deviding by zero is not allowed. So where does this equation fit in the phrase quoted from the book. I am not an expert just trying to understand some – Asim Jun 03 '20 at 08:07
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    The book is pointing out Euler's mistake in using values of $x$ outside the domain of convergence. It is not stating the equation is correct. Quite the opposite, actually. – superckl Jun 03 '20 at 08:48
  • @superckl thanks man. R u sure though? – Asim Jun 03 '20 at 08:49
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    @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. – superckl Jun 03 '20 at 08:51
  • This is why i am here for. English is not my language. I am just trying to figure out what actually book is saying. – Asim Jun 03 '20 at 08:55
  • @superckl You are the only one who ACTUALLY answered my question. Thanks a lot man. – Asim Jun 03 '20 at 09:18
  • Euler should have used the value for $|x|\ge 1$ where the sum does not converge ? Hard to believe ! Maybe, Euler wanted to make a joke. This is similar to the "equation" $1+2+3+\cdots =-1/12$ that is defended on this site by some users like a religion. – Peter Jun 03 '20 at 12:30

2 Answers2

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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

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

Jan
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  • What exactly arises your confusion? – Jan Jun 03 '20 at 07:37
  • 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 ? – Asim Jun 03 '20 at 07:43
  • 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: https://math.stackexchange.com/questions/29023/values-of-sum-n-0-infty-xn-and-sum-n-0n-xn – Jan Jun 03 '20 at 07:46
  • 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. – Asim Jun 03 '20 at 07:51
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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, $|x|≠ 1$.

LL 3.14
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