Question Number 2720 by prakash jain last updated on 25/Nov/15

Analytical Continuation Sum of the below divergent series was shown to be using analytical continuation. Σ_(i=1) ^n 2^(i−1) =−1 ...(A) ζ(−1)=Σ_(i=0) ^∞ i=−(1/(12)) ...(B) While reading about analytical continuation, I found the found the following: If f_1 is a analytic function over domain D_1 and If f_2 is a analytic function over domain D_2 and f_1 =f_2 on D_1 ∩D_2 , f_2 is called analytical continuation of f_1 and vice versa. Question: In case of series (A) and (B) above what function is used as f_2 to find the sum also what is used as f_1 ?

Commented byprakash jain last updated on 25/Nov/15

There is also a question below by Filup about definition of ζ(s) for s<1 using analytical continuity.

Answered by 123456 last updated on 26/Nov/15

for A f_1 (x)=Σ_(i=0) ^(+∞) x^i f_2 (x)=(1/(1−x)) f_1 (x)=lim_(n→+∞) Σ_(i=0) ^n x^i =lim_(n→+∞) ((x^(n+1) −1)/(x−1)) ∣x∣<1⇒x^(n+1) →0⇒f_1 (x)=((0−1)/(x−1))=(1/(1−x)) f_1 (x)=f_2 (x) ∣x∣<1, so f_2 (x)=(1/(1−x)) is a analytic continuation

Commented by123456 last updated on 26/Nov/15

S=Σ_(i=0) ^n x^i xS=xΣ_(i=0) ^n x^i =Σ_(i=0) ^n x^(i+1) =Σ_(i=1) ^(n+1) x^i (x−1)S=Σ_(i=1) ^(n+1) x^i −Σ_(i=0) ^n x^i (x−1)S=x^(n+1) −1 S=((x^(n+1) −1)/(x−1))

Commented byprakash jain last updated on 26/Nov/15

Thank You.