Abel Summation

Calculus Level 2

1 − 2 + 3 − 4 + 5 − 6 + ⋯ 1-2+3-4+5-6+\cdots

Define the Abel sum ∑ n = 0 ∞ a n \sum\limits_{n=0}^\infty a_n to be lim ⁡ z → 1 − ∑ n = 0 ∞ a n z n , \displaystyle \lim_{z\to 1^-} \sum_{n=0}^\infty a_nz^n, if that limit exists.

The Abel sum of the (divergent) series as shown above can be written as a b \frac{a}{b} , where a a and b b are coprime positive integers. Find a + b a+b .


The answer is 5.

View wiki
Patrick Corn by Patrick Corn , 44, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

展豪 張
Mar 8, 2016

The following is my (stupid) approach:
f ( z ) = 1 − 2 z + 3 z 2 − 4 z 3 + . . . f(z)=1-2z+3z^2-4z^3+...
∫ f ( z ) d z = C + z − z 2 + z 3 − z 4 + . . . \int f(z) dz = C+z-z^2+z^3-z^4+...
= C + z 1 + z =C+\frac z{1+z}
f ( z ) = d d z ( C + z 1 + z ) f(z)=\frac d{dz} (C+\frac z{1+z})
= 1 ( 1 + z ) 2 =\frac 1{(1+z)^2}
lim ⁡ z → 1 − f ( z ) = 1 4 \displaystyle\lim_{z\to 1^-}f(z)=\frac 14


11 Helpful 4 Interesting 8 Brilliant 0 Confused
Otto Bretscher
Feb 29, 2016

We have the arithmetic-geometric series ,

∑ k = 1 ∞ ( − 1 ) k + 1 k z k = z ( z + 1 ) 2 , \large \sum_{k=1}^{\infty}(-1)^{k+1}k \; z^k=\dfrac{z}{(z+1)^2}, with limit 1 4 \frac{1}{4} . The answer is 5 \boxed{5}

11 Helpful 1 Interesting 2 Brilliant 0 Confused
Bharath Sriraam
Feb 29, 2016

L e t S = 1 − 1 + 1 − 1 + . . . S = 1 − 1 + 1 − . . . A d d i n g t h e 2 e q u a t i o n s w e g e t , 2 S = 1 S o S = 1 / 2 L e t B = 1 − 2 + 3 − 4 + 5 − . . . B = 1 − 2 + 3 − 4 + . . . A d d i n g t h e 2 e q u a t i o n s w e g e t , 2 B = 1 − 1 + 1 − 1 + . . . S o 2 B = S B u t S = 1 / 2 S o B = 1 / 4 H e n c e t h e r e q u i r e d v a l u e i s 1 + 4 = 5. Let\quad S\quad =\quad 1-1+1-1+...\\ \quad \quad \quad S\quad =\quad \quad \quad 1-1+1-...\\ Adding\quad the\quad 2\quad equations\quad we\quad get,\\ \quad \quad 2S\quad =\quad 1\\ So\quad S\quad =\quad 1/2\\ \\ Let\quad B\quad =\quad 1-2+3-4+5-...\\ \quad \quad \quad B\quad =\quad \quad \quad 1-2+3-4+...\\ Adding\quad the\quad 2\quad equations\quad we\quad get,\\ \quad \quad 2B\quad =\quad 1-1+1-1+...\\ So\quad 2B\quad =\quad S\\ But\quad S\quad =\quad 1/2\\ So\quad B\quad =\quad 1/4\\ Hence\quad the\quad required\quad value\quad is\quad 1+4=5.\\

9 Helpful 0 Interesting 3 Brilliant 0 Confused

This doesn't answer the question of "What is the Abel sum".

You are using a different system to compute the value, and should be explicit about what "=" means in your equations.

Calvin Lin Staff - 5 years, 3 months ago

Log in to reply

I am sorry sir but the approach I used is the one used in String theory to prove that the sum of natural numbers till infinity is -1/12. I just used their approach to get the sum of the series mentioned in the beginning of the question.

Pardon me but I'm new with Latex so I'm not used to it. Thank you.

Bharath Sriraam - 5 years, 3 months ago

Log in to reply

My question is "How do you know that the Abel sum approach, yields the same answer as the string theory approach"? E.g. If we the "Abel_2" sum was lim ⁡ ∑ a n z 2 n + 3 \lim \sum a_n z^{2n+3} , must the answer be the same as some other approach that you chose?

Note: Your solution, rigorously speaking, isn't exactly the "string theory approach". It is a simplification which explains how they came to that value initially, but then also had to back it up with further understanding.

Calvin Lin Staff - 5 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...