Modulus of a Sequence: How to Prove Nullness?

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The discussion focuses on proving the nullness of the sequence defined as {(1+(-1)^n)/(n+(-1)^n)}. The method outlined in the referenced textbook requires demonstrating that for every ε>0, there exists an integer N such that the modulus of the sequence is less than ε for all n>N. The modulus of the sequence is evaluated as |(1+(-1)^n)/(n+(-1)^n)|, which simplifies to 2/(n+1) for even n and 0 for odd n. The sequence begins at n=2, confirming that negative values of n are irrelevant in this context.

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rohan03
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I know in general how to prove if sequence is null or not . But here is my confusion
- method in the textbook I am reading - asks that to prove that any sequence is null we must show that for each ε>0, there is an integer N such that modulus of the given sequence is < ε, for all n>N

now I also understand this but my dilema is
taking modulu say for example given sequence is
{(1+(-1)^n)/(n+(-1)^n )}
taking modulus gives:
|(1+(-1)^n)/(n+(-1)^n )| is modulus 2/(n+1) ?

just not sure
 
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rohan03 said:
I know in general how to prove if sequence is null or not . But here is my confusion
- method in the textbook I am reading - asks that to prove that any sequence is null we must show that for each ε>0, there is an integer N such that modulus of the given sequence is < ε, for all n>N

now I also understand this but my dilema is
taking modulu say for example given sequence is
{(1+(-1)^n)/(n+(-1)^n )}
taking modulus gives:
|(1+(-1)^n)/(n+(-1)^n )| is modulus 2/(n+1) ?

just not sure


\left|\frac{1+(-1)^n}{n+(-1)^n}\right|=\left\{\begin{array}{cc}0\,&amp;\,if\,\,n\,\,is\,\,odd\\\frac{2}{n+1}\,&amp;\,if\,\,n\,\,is\,\,even\end{array}\right. , and of course the seq. begin with n=2.

DonAntonio

Ps. Of course, the absolute value has no relevance here...did you notice this?
 
Yes the question states n=2,3... So I am correct and must ignore negative values of n
 

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