Proving Divergence of ((-1)^m)m

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SUMMARY

The sequence ((-1)^m)m diverges, as established through the epsilon-delta definition of convergence. By assuming convergence and setting epsilon to 1/2, it is shown that for sufficiently large m (specifically m = 2M), the terms of the sequence do not satisfy the convergence condition. The analysis confirms that the absolute difference between the sequence terms and any limit L cannot remain bounded by epsilon, leading to the conclusion of divergence.

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Homework Statement


Prove sequence ((-1)^m)m diverges.

Homework Equations



for all epsilon greater than zero, there exists a natural number M such that for all natural numbers m greater than or equal to M, Ix_m-xI is less than or equal to epsilon.[/B]

The Attempt at a Solution



Assume it converges and consider epsilon to be 1/2, so there is an M such that I((-1)^m)m-LI is less than or equal to epsilon for all m greater than or equal to M. Now consider m to be 2M, Why is m greater than or equal to 1/2 for when m=2M? Just help me with this part, and not anything else, please.
 
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Mathematicsresear said:
Why is m greater than or equal to 1/2 for when m=2M?

Did you mean to ask why ##(-1)^m m \ge 1/2## when ## m = 2M##?
 
Stephen Tashi said:
Did you mean to ask why ##(-1)^m m \ge 1/2## when ## m = 2M##?
no
 

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