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I need to find the limit point for the following set

$A = \left( (-1)^{m} + \dfrac{(-1)^n}{n}\right) \text{m,n} \in \mathbb N$

So, Here is what I did , If I fix $m$ and vary $n$ then as $n \to \infty$ clearly $+1,-1$ are two limit points.

Also If I fix $n$ and vary $m$ then $\left( 1 + \dfrac{(-1)^n}{n}\right)$ and $\left( -1 + \dfrac{(-1)^n}{n}\right)$ are other two limit points.

Is this correct ?

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1 Answer 1

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If we refer to $m,n\to \infty$ then since $\dfrac{(-1)^n}{n}\to 0$ all boils down in $(-1)^{m}$ then limit points are $1$ and $-1$, otherwise also the other two limit points exists and they are dependent upon $n$.

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  • $\begingroup$ But In the second case I have fixed a value $n \in N$. Then why are you taking taking the limit $n \to \infty$ ? Can you please explain this. $\endgroup$
    – zeroflank
    Oct 25, 2019 at 17:32
  • $\begingroup$ @zeroflank If we refer to $m,n\to \infty$ then limit points are just $\pm1$ otherwise yes you are correct there are also other two limit points depending upon $n$. $\endgroup$
    – user
    Oct 25, 2019 at 17:36
  • $\begingroup$ Thank you, for such a quick response. $\endgroup$
    – zeroflank
    Oct 25, 2019 at 17:41

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