Inspired by Pi Han Goh, Part 2

Calculus Level 5

$1-2+3-4+...$

The series above diverges, of course. Its Abel sum is defined as

$A=\lim_{x\to 1^-}\sum_{n=0}^{\infty}(-1)^{n+1} n x^n$

Find $A$ to three significant figures. Enter 666 if you come to the conclusion that no such Abel sum $A$ exists.

Twin Problem

The answer is 0.25.

4 solutions

Chew-Seong Cheong
May 16, 2016

$\begin{aligned} A & = \lim_{x \to 1^-} \sum_{n=0}^\infty (-1)^{n+1}nx^n \\ & = \lim_{x \to 1^-} \sum_{n=0}^\infty (-1)^{n+1}x \frac{d}{dx} x^n \\ & = \lim_{x \to 1^-} x \frac{d}{dx} \sum_{n=0}^\infty (-1)^{n+1} x^n \\ & = \lim_{x \to 1^-} x \frac{d}{dx} \left(-\frac{1}{1+x} \right) \quad \text{for }|x| < 1 \\ & = \lim_{x \to 1^-} \frac{x}{(1+x)^2} \\ & = \frac{1}{4} = \boxed{0.25} \end{aligned}$

Using Maclaurin expansion, we have $\displaystyle \frac{1}{(1+x)^2} = \sum_{n=0}^\infty (-1)^{n+1}\color{#D61F06}{n!} x^n$ .

Chew-Seong Cheong - 5 years, 1 month ago

I don't get this one. Is that a factorial in the power series?!

Otto Bretscher - 5 years, 1 month ago

Hana's solution of $\displaystyle \frac{1}{(1+x)^2}$ is incorrect.

Chew-Seong Cheong - 5 years, 1 month ago

Hana factors out an $x$ ... her formula is correct.

I'm not so sure about your formula $\frac{1}{(1+x)^2}=\sum_{n=0}^{\infty}(-1)^{n+1}n!x^n$ , though ;)

Otto Bretscher - 5 years, 1 month ago

Yes, very nice, Comrade! (+1) This is analogous to Hana's solution but written out in more detail. Now, can you apply this technique here ? ;)

Otto Bretscher - 5 years, 1 month ago

Hana Wehbi
May 15, 2016

$1-2x+3x^{2}-4x^{3}+...=$ $\frac{1}{(1+x)^{2}}$ = $\frac{1}{4}$ as $x$ goes to 1.

Thus, A= $lim_{x\to1^{-}}$ $\frac{1}{(1+x)^{2}}$ = $\frac{1}{4}$ =0.25

I am just trying to add to other solutions, hopefully, I came up with something impressive. If not, be easy with me :)

I took the $\sum_{n=1}^{infinity}$ $n(-x)^{n-1}$

Exactly! (+1) Your solution is short, simple, and elegant... impressive indeed! Thank you!

Otto Bretscher - 5 years, 1 month ago

Thank you.

Hana Wehbi - 5 years, 1 month ago

Dragan Marković
May 15, 2016

$1-1+1-1+...=S$ $1-S=1-(1-1+1-1+..)$ $1-S=S$ $S= \frac{1}{2}$ Now let $1-2+3-4+...=S_{1}$ $S_{1}+ S_{1} = 1-2+3-4+..$ By shifting it up $1-2+3-4+...$ $2S_{1}=1-1+1-1+... 2S_{1}=\frac{1}{2} => S_{1}=\frac{1}{4}=0.25$

That's how Euler did it. But this deduction does not live up to modern standards of rigor in analysis. We know that $1-1+1-1_...$ diverges, so , the equation $1-1+1-1+...=S$ is meaningless, although it does (mysteriously) produce the right answer. Euler himself was very puzzled by this result when he wrote in 1749: "...when it is said that the sum of this series 1−2+3−4+5−6 etc. is 1⁄4, that must appear paradoxical. For by adding 100 terms of this series, we get −50, however, the sum of 101 terms gives +51, which is quite different from 1⁄4 and becomes still greater when one increases the number of terms. But I have already noticed at a previous time, that it is necessary to give to the word sum a more extended meaning." This "extended meaning" was later introduced by Abel.

Otto Bretscher - 5 years, 1 month ago

Oh well this is the way howI was taught to estimate the $1-2+3-4+...$

Dragan Marković - 5 years, 1 month ago

Well, you did end up with the right answer ;)

Otto Bretscher - 5 years, 1 month ago

Sir basically, this is an airthmatic geometric mean that we calculated, isn't it?