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There is the following equation in a text I am reading

$$ \sum_{n=0}^\infty (1 - x) x^k = 1. $$

I would like to know where does the identitity come from.

Is it from some telescopic property, e.g., \begin{align} \sum_{k=0}^\infty(1 - x) x^k &= \sum_{k=0}^\infty(1 - x) x^k\\ &= \sum_{k=0}^\infty x^k - x^{k+1}\\ &= [(x^0 - x^1) + (x^1 - x^2)] \sum_{k=2}^\infty x^k - x^{k+1}\\ &= [(x^0 - x^2)] \sum_{k=2}^\infty x^k - x^{k+1}\\ &= 1 - x^2 \sum_{k=2}^\infty x^k - x^{k+1}, \end{align}

which, in the limit $x \to \infty$, would render the sum 0?

As can be seen from @qbert's answer, this seems to be a particular case of Value of $\sum\limits_n x^n$, where we want to calculate

$$(1-x)\sum_n x^n,$$

for $|x| < 1$.

Rafael Beirigo
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2 Answers2

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It is just the geometric series identity $$ \sum_{k=0}^\infty\lambda^k=\frac{1}{1-\lambda} $$ So, $$ (1-\lambda)\sum_{k=0}^\infty x\lambda^k\\ =x\left(\sum_{k=0}^\infty\lambda^k-\sum_{k=0}^\infty\lambda^{k+1}\right)\\ =x\left(\frac{1}{1-\lambda}-\lambda \sum_{k=0}^\infty\lambda^{k}\right)\\ =x\left(\frac{1}{1-\lambda}-\lambda \frac{1}{1-\lambda}\right)\\ =x\ $$ where we used linearity of summation a bunch.

operatorerror
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You used telescoping in a wrong way.

$$\sum_{k=0}^n(x^k-x^{k+1})=x^0-x^1+x^1-x^2+\cdots-x^{n+1}=1-x^{n+1},$$

so that

$$\sum_{k=0}^\infty(x^k-x^{k+1})=\lim_{n\to\infty}\sum_{k=0}^n(x^k-x^{k+1})=1$$ when $|x|<1.$