Can I Cancel Them In Pairs?

Calculus Level 3

$\large \lim _{ x\to -1^-} \left( \sum _{n=0}^\infty x^{-n} \right) =\, ?$

-1

-\frac { 1 }{ 2 }

0

\frac { 1 }{ 2 }

1

Limit does not exist

1 solution

Chew-Seong Cheong
Oct 7, 2016

Relevant wiki: Geometric Progression Sum

Recall that the infinite geometric progression sum with first term $a$ and common ratio $r$ can be expressed as $\frac {a}{1-r}$ , where $-1<r<1$ . The series in question is an infinite geometric progression sum.

$\begin{aligned} L & = \lim_{x \to -1^-} \sum_{n=0}^\infty x^{-n} \\ & = \lim_{x \to -1^-} \frac 1{1-\frac 1x} & \text{for } -1 < \frac 1x < 1 \\ & = \lim_{ \color{#D61F06}{x \to -1^-}} \frac 1{1-\frac 1x} & \text{for } \color{#D61F06}{x < -1}, x > 1 \\ & = \boxed{\dfrac 12} \end{aligned}$

I can imagine substituting the value of $x$ with $-1$ to get the value that comes from Ramanujan Summation . Very sweet problem you have here, even Wolfram couldn't even give the right answer for the problem. ^.^

Michael Huang - 4 years, 7 months ago

Can I Cancel Them In Pairs?

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