Is my proof of this sequence's divergence good enough?

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    Divergence Proof
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SUMMARY

The series Ʃ n=1 to infinity of cos(n∏) diverges as established by the nth term test. The sequence defined by an=cos(n∏) oscillates between -1 and 1, resulting in a limit that does not approach zero. Therefore, the series fails to meet the necessary condition for convergence, confirming its divergence. The proof presented is fundamentally sound, although it could benefit from more precise mathematical language.

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Ʃ n=1 to infinity of cos(n∏)
letting an=cos(n∏), I rewrote this as (-1)^n=an.

Using the nth term test i let the limit as n->∞ go to infinity. This value bounces back and forth between positive and negative, but I know clearly the value =/= 0, therefore it diverges by the nth term test.

Is there anything I should add to my proof?
 
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mathnoobie said:
Ʃ n=1 to infinity of cos(n∏)
letting an=cos(n∏), I rewrote this as (-1)^n=an.

Using the nth term test i let the limit as n->∞ go to infinity. This value bounces back and forth between positive and negative, but I know clearly the value =/= 0, therefore it diverges by the nth term test.

Is there anything I should add to my proof?



Looks fine to me, although with a somewhat folkloric language as "bouncing back and forth" and "n-th term test", which I don't what

is. Perhaps you meant simply that the series doesn't converge as its general element's sequence doesn't converge to zero, which is a necessary condition.

DonAntonio
 

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