My answer: The limit of the above sequence doesn't exist. And what about limit points?? The answer shows $-1\; and \;1$ But definition says if there exist any point in every nbd of $1$ except $1$ then $1 $ is a limit point of the sequence but sadly I don't find any no.
Please Help.
Make sure you are correctly understanding the definition of a accumulation point, sometimes also called a limit point, and how it differs from a limit.
In informal terms - the limit of a sequence, if it exists, is some value which all the points in the sequence eventually get arbitrarily close to, as the sequence progresses.
The sequence defined by $a_n = (-1)^n$ looks like this: $[1, -1, 1, -1, 1, -1, ...]$, that is assuming $n$ starts at $0$. Since the sequence keeps oscillating between $1$ and $-1$, it never gets "close" to any point, and thus it does not have a limit.
A accumulation point of a sequence - of which, by the way, there can be more than one! - is any point which the sequence "revisits" infinitely often. However, "revisits", doesn't necessarily mean it has to touch that point, although it can. It just has to get close, and there has to be some time further along when it gets at least as close if not closer.
So for your sequence, its accumulation points are $1$ and $-1$, as no matter how far you go down the sequence, it keeps hitting them (right on the nose, as it turns out) again & again.
By the way - this is why, if a sequence has some accumulation point, then that means there is a subsequence with that accumulation point as its limit. Take the given example. You can take the subsequence of the 0th, 2nd, 4th, etc. terms which is $[1, 1, 1, ...]$. Clearly this has a limit of $1$ since it's literally just a sequence of $1$'s. The same is true with the 1st, 3rd, 5th, etc. terms and $-1$.
Just for fun... here's another sequence to consider, starting at $n=0$: $$ a = \left[ \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{3}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{1}{5}, \frac{1}{5}, \frac{4}{5}, \frac{1}{6}, \frac{1}{6}, \frac{5}{6}, ... \right] $$
Its limit does not exist. This is because every third term, it is alternating, ad infinitum.
However it does have two accumulation points: $0$ and $1$. Here are the subsequences which converge to each of these accumulation points, respectively: $$ b = \left[ \frac{1}{2}, \frac{1}{2}, \text{ }\text{ }\text{ } \frac{1}{3}, \frac{1}{3}, \text{ }\text{ }\text{ } \frac{1}{4}, \frac{1}{4}, \text{ }\text{ }\text{ } \frac{1}{5}, \frac{1}{5}, \text{ }\text{ }\text{ } \frac{1}{6}, \frac{1}{6}, \text{ }\text{ }\text{ } ... \right] \rightarrow 0 \\ c = \left[ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \frac{1}{2}, \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \frac{2}{3}, \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \frac{3}{4}, \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \frac{4}{5}, \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \frac{5}{6}, ... \right] \rightarrow 1 $$
Note that it does not matter that the "small" fractions repeat while the "big" ones don't; the accumulation points are still $0$ and $1$.
The "in every nbd of 1 except {1}" part just means we can't demand that this nbd be so small as to be the singleton set {1} itself. The nbd has to have some interior to it, though it could be arbitrarily small.
Thus, the sequence need not actually ever touch 1, for 1 to still be a limit point. It just has to "revisit" 1 infinitely closely, infinitely often.
– indnwkybrd Apr 14 '23 at 01:46"Hey, $1$... we only expected you to get arbitrarily close to that sequence, and we still woulda called you one of its limit points... but, hey, if ya wanna go ahead & touch it? Then even better!"
– indnwkybrd Apr 14 '23 at 01:49