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Incidentally, I found

$$\frac{1}{\sum_{n=1} \frac{1}{x^{n}}} = (x-1)$$

where $x\ge 2$.

Please direct me to how others have developed the relationship. My computer cannot compute more than X = 500,000 so I just infer that it is always true.

926reals
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  • This is just a geometric series. Sorry, Im so vague for a long explanation, but you can find a lot of information about it here and in the net. – Masacroso Jul 22 '16 at 04:47
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    How did you get that? I'm getting $\sum_{n=1}^\infty \frac{1}{x^{n}}=\frac{1}{1-\frac{1}{x}}-1=\frac{x}{x-1}-1=\frac{1}{x-1}$ – Aweygan Jul 22 '16 at 04:48
  • It's not true. If y>x then $1/y^n < 1/x^n $ so \sum 1/y^n < \sum 1/x^n$ but y-1 > x-1. Not compatible. – fleablood Jul 22 '16 at 04:55
  • $(x-1)\sum_{n=1}^m 1/x^n = \sum_{n=1} 1/x^{n-1} -\sum_{n=1}^m 1/x^n=\sum_{n=0}^{m-1} 1/x^n - \sum_{n=1}^m 1/x^n = 1 - 1/x^m $. So SUMto_m = (1-1/x^m)/(x-1). If |x|>1 then inf sum = limit SUMto_m = 1/(x-1). – fleablood Jul 22 '16 at 05:11
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    (-1) because I found this question on the Reopen queue after a pointless edit. –  Jul 22 '16 at 06:59

2 Answers2

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It is not true. It is a geometric series and for $x \gt 1$ we have $$\sum_{n=1}^\infty \frac{1}{x^{n}} = \frac1{x-1}$$ For $x$ large, the sum is clearly less than $1$, while your expression has the right side much larger than $1$

The question is changed, and is now equivalent to the above. The informal way to see it is $$S=\frac 1x + \frac 1{x^2}+\frac 1{x^3}+\ldots\\ xS=1+\frac 1x + \frac 1{x^2}+\frac 1{x^3}+\ldots\\ xS-S=1\\S=\frac 1{x-1}$$ More formally, you need to develop the machinery of a geometric series. The equation is correct for all real $x \gt 1$

Ross Millikan
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  • Maybe it's too late at night for me, but if the relation in the OP holds, why am I getting $$\dfrac{1}{\sum_{i=1}^n \frac{1}{x^{i}}} = (x-1)$$

    $$1 = \sum_{i=1}^n \frac{1}{x^{i}} (x-1)$$

    $$1 = \sum_{i=1}^n \frac{1}{x^{i-1}} - \sum_{i=1}^n \frac{1}{x^{i}} $$

    $$1 = (1 + \dfrac {1}{x} + \cdots + \dfrac {1}{x^{n-1}}) - (\dfrac {1}{x} + \cdots + \dfrac {1}{x^{n}})$$

    $$1 = 1-\dfrac {1}{x^n}$$

     – Ovi Jul 22 '16 at 06:34
  • @Ross Your first equality is true for the sum beginning at $;n=0;$, not at $;n=1;$ , and it is true then for general $;|x|>1;$ – DonAntonio Jul 22 '16 at 08:02
  • @DonAntonio: No, it is correct starting at $n=1$. For $x=2$ it reads $1=\frac 12 + \frac 14 +\frac 18+\ldots$, which is correct. An added $1$ on the right from the $n=0$ term would spoil it – Ross Millikan Jul 22 '16 at 13:25
  • @RossMillikan You are completely correct. I missed the fact that the quotient here is $;\frac1x;$ , not $;x;$ . – DonAntonio Jul 22 '16 at 13:47
  • Oh I found the mistake after a good night's sleep, the series in your answer goes up to $\infty$, not $n$ – Ovi Jul 22 '16 at 14:05
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A series is defined as

$$\sum_{k=c}^{\infty}b_k=\lim_{n\to \infty}\sum_{k=c}^{n}b_k$$

And we can define a partial geometric sum as

$$S=\sum_{k=0}^{n}ax^k=a\sum_{k=0}^n x^k$$

To see the relationship of your question we must notice that

$$S=xS+a-ax^{n+1}\to S-xS=a(1-x^{n+1})\to S=a\frac{1-x^{n+1}}{1-x}$$

Then we have the relationship

$$S=\sum_{k=0}^n ax^k=a\frac{1-x^{n+1}}{1-x}$$

Then the value of the series

$$\sum_{k=0}^\infty ax^k=\lim_{n\to \infty} S=\lim_{n\to\infty}a\frac{1-x^{n+1}}{1-x}=\begin{cases}+\infty\cdot\text{sgn}(a) &,\text{when }x\ge 1\\\text{undefined}&,\text{when }x\le -1\\\frac{a}{1-x} &,\text{when }x\in(-1,1)\end{cases}$$

Then the geometric series only converges when $|x|<1$. And for your case notice that

$$\sum_{k=1}^\infty ax^k=-a+\sum_{k=0}^\infty ax^k$$

Masacroso
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