-1

The binomial expansion of $(a+b)^{-2}$ is given as

$$(a+b)^{-2}=\sum_{n=1}^\infty(-1)^{n+1}na^{-1-n}b^{n-1}\tag{I think}$$

And when $a=b=1$,

$$2^{-2}=\sum_{n=1}^\infty(-1)^{n+1}n=1-2+3-4+\dots$$

So I was wondering if this were a way to evaluate the divergent summation in a ramanujan sort of meaning.

Simply Beautiful Art
  • 76,603
  • 2
    you should study a complex analysis course – reuns Apr 20 '16 at 00:35
  • @user1952009 sadly, that is difficult without having first taken Calculus, and I'm a high school student, self taught everything I know on this site. :/ More self teaching is difficult. – Simply Beautiful Art Apr 20 '16 at 16:58
  • no, I know your level and I tell you you are capable. of course you'll need re-proving first some calculus theorems about integrals derivatives sequences series and trigonometric functions, but you can also prove those during studying a complex analysis course (and you'll have all your divergent series examples to help you). – reuns Apr 20 '16 at 17:16
  • @user1952009 Cool, thanks for the confidence boost, but darn, with my actually classes, the amount of time I get to digest new math material is shrinking. – Simply Beautiful Art Apr 20 '16 at 17:19
  • @user1952009 I took your advice and learned some complex analysis :D – Simply Beautiful Art Feb 15 '17 at 14:17
  • @reuns So I did some complex analysis. What else would you recommend? – Simply Beautiful Art Aug 30 '17 at 19:56
  • Hi. I looked at your questions and answers I didn't find many Fourier analysis topics. See thisone about proving the Fourier inversion theorem in a few lines. 1 year ago I didn't know anything about algebraic number fields and abstract algebra, and after reading questions MSE and wikipedia I could obtain a basic knowledge for studying this topic. – reuns Aug 30 '17 at 20:10
  • Cool, thanks :-) @reuns – Simply Beautiful Art Aug 30 '17 at 20:12

1 Answers1

1

Yes this is Abel Summation (Regularization) because it comes from defining:

$$f(z)=\sum_{n=0}^\infty a_nz^n$$ and evaluating:

$$A:=\lim_{z\rightarrow 1^{-}}\sum_{n=0}^\infty a_nz^{n}.$$

In your case:

$$f(z)=\frac{1}{(1+z)^2}=\sum_{n=1}^\infty ((-1)^nn)z^n.$$

You can find other ways to regularize your sum here.

Alex R.
  • 33,289