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How would I go about finding all accumulation points (or limit points?) of a sequence? I know that I need to find subsequences that converge to that point. But for that, I first need to find that point. How do I do this? And how can I be sure that I found all accumulation points?


The specific exercise is:

Find all accumulation points of the following sequence:

$(1+ (-1)^n) \frac {n + 1} n + (-1)^n$


Any help, especially for the example above, would be appreciated.

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  • $\begingroup$ Consider the cases when $n$ odd/even separately $\endgroup$
    – who
    Oct 25 '19 at 9:43
  • $\begingroup$ As a general tip: plot the first some values and try to find a candidate. Then show that this candidate works $\endgroup$ Oct 25 '19 at 9:43
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For $n$ odd you get just $-1$. For $n$ even you get the sequence $\frac {2(n+1)} n +1$ which converges to $3$. So what are the limit points?

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The example makes it easier by basically dividing the naturals into two: even and odd. In fact, for any odd number you have that your sequence equals $-1$, so it is trivially convergent to it.

For even $n$ instead you get: $$ 2\frac{n+1}{n} + 1$$ Which means it converges to $3$.

Now to show that these are the only ones, you can just notice that any way of "skipping from natural number to natural number" that isn't strictly just made of only odds or only evens from a certain point on, will continue to switch between trying to converge to $3$ and trying to converge to $-1$, hence not converging to anything in particular.

EDIT: Solving these problems in general is very hard, especially when the pattern isn't clear. A trick you can use is exactly to reason about known patters: multiples of $2$, or $3$ or any number, prime numbers, perfect squares and so on. The idea is to pin down enough convergent subsequences to cover all the points of $\Bbb{N}$.

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The sequence is not convergent but $3$ is the only limit point. You can see for large $n's$ there are infinitely many points of sequence is near $3$.

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