Analytical-Continuation-Sum-of-the-below-divergent-series-was-shown-to-be-using-analytical-continuation-i-1-n-2-i-1-1-A-1-i-0-i-1-12-B-While-reading-about

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 .

\underset{i = 1}{\sum^{n}} 2^{i - 1} = - 1 \dots (A)

ζ (- 1) = \underset{i = 0}{\sum^{\infty}} i = - \frac{1}{12} \dots (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} \cap 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 by prakash 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) = \underset{i = 0}{\sum^{+ \infty}} x^{i}

f_{2} (x) = \frac{1}{1 - x}

f_{1} (x) = \lim_{n \to + \infty} \underset{i = 0}{\sum^{n}} x^{i} = \lim_{n \to + \infty} \frac{x^{n + 1} - 1}{x - 1}

∣ x ∣< 1 \Rightarrow x^{n + 1} \to 0 \Rightarrow f_{1} (x) = \frac{0 - 1}{x - 1} = \frac{1}{1 - x}

f_{1} (x) = f_{2} (x) ∣ x ∣< 1, so f_{2} (x) = \frac{1}{1 - x} is a analytic continuation

Commented by 123456 last updated on 26/Nov/15

S = \underset{i = 0}{\sum^{n}} x^{i}

x S = x \underset{i = 0}{\sum^{n}} x^{i} = \underset{i = 0}{\sum^{n}} x^{i + 1} = \underset{i = 1}{\sum^{n + 1}} x^{i}

(x - 1) S = \underset{i = 1}{\sum^{n + 1}} x^{i} - \underset{i = 0}{\sum^{n}} x^{i}

(x - 1) S = x^{n + 1} - 1

S = \frac{x^{n + 1} - 1}{x - 1}

Commented by prakash jain last updated on 26/Nov/15

Thank You .