Question

Prove that ${\bf \zeta ( - 1 ) = - \dfrac{1}{12}}$​

Answer 1

The zeta function is defined as:

$\boxed{\displaystyle\zeta(s) = \sum\limits_{i = 1}^\infty \dfrac{1}{i^s}}$

By substituting the value of s = -1, we get:

$\displaystyle\implies \zeta(-1) = \dfrac{1}{1^{-1}} + \dfrac{1}{2^{-1}} + \dfrac{1}{3^{-1}}+ ...$

$\displaystyle\implies \zeta(-1) = 1 + 2 + 3 + ...$

Now this is very well known Ramanujan's Paradox that the sum of all the natural numbers is -1/12. Let's prove it!!

Consider a series:

A = 1 - 1 + 1 - 1 + 1 - 1 +...

A = 1 -(1 - 1 + 1 - 1 + ....)

A = 1 - A

2A = 1

A = 1/2

Consider another series:

B = 1 - 2 + 3 - 4 + 5 - 6 +...

Now perform an operation A - B

A - B = (1 - 1 + 1 - 1 + ... ) - (1 -2 + 3 - 4 + ...)

A - B = (1 - 1) + (-1 + 2) + (1 - 3) + (-1 + 4) + ...

A - B = 0 + 1 - 2 + 3 - 4 + ...

A - B = B

A = 2B

1/2 = 2B

B = 1/4

Consider our required series:

S = 1 + 2 + 3 + 4 + ...

Now perform an operation B - S

B - S = (1 - 2 + 3 - 4 + ...) - (1 + 2 + 3 + 4 + ...)

B - S = (1 - 1) + (-2 -2) + (3 -3) + (-4 - 4) +...

B - S = 0 - 4 + 0 - 8 + ...

B - S = -4 - 8 - 12 +...

B - S = -4( 1  + 2 + 3 + 4 + ...)

B - S = -4(S)

B - S = -4S

3S = - B

3S = - 1/4

S = -1/12

Hence the required result is proved that:

$\underline{\boxed{\zeta(-1) = -\dfrac{1}{12}}}$

Answer 2

If ${\bf {\sigma}_1}$ , ${\bf {\sigma}_2}$ , ${\bf \cdots \cdots}$ , ${\bf {\sigma}_n}$ are distinct automorphism of ${\mathbb K}$ , then show that it is impossible to find elements ${\bf a_1}$ , ${\bf a_2}$ , ${\bf \cdots \cdots}$ , ${\bf a_n}$ not all zero in ${\mathbb K}$ such that ;

${\boxed{\bf a_{1} \sigma_{1} ( u ) + a_{2} \sigma_{2} + \cdots \cdots + a_{n} \sigma_{n} ( u ) = 0 \quad \forall \bf u \in \mathbb K}}$

Topic :- Advance Abstract Algebra MSc 1st sem.

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