Consider the sequence $$ (x_n)_{n \in \mathbb{N}} = \frac{n}{n+1} + \sin \frac{n\pi}{2} $$ and calculate its limit points, as well as the $\lim \sup_{n \to \infty} x_n$ and $\lim \inf_{n \to \infty} x_n$.

My first approach here would be to look at the limits of the elements of $x_n$.

For $y_n := \frac{n}{n+1}$, the limit is obviously $\lim_{n \to \infty} y_n = 1$. $z_n := \sin \frac{n\pi}{2}$ desn't have a definitive limit, but rather the limit points $-1,0,1$ as per the definition that infinitely many elements of the sequene are arbitrarily close to one of these values (rather obvious as well).

But: can I say that because the "subsequences" of $x_n$ have these properties, the limit points of $x_n$ are $0,1,2$? Without further manipulation of $x_n$ as a whole (not just for this sequence, but in general for the addition/subtraction of sequences)? This is Calculus 1, so the concepts are still rather basic.

As for the $\lim\sup$ and $\lim\inf$, my intuition would be $\lim\sup_{n \to \infty} x_n = 2$ and $\lim\inf_{n \to \infty} x_n = 0$, based on similar approaches as for the limit points.

Any help would be much appreciated :)


We have $$\lim_{n\to+\infty}x_{2n}=\lim_{n\to+\infty}(\frac{2n}{2n+1}+\sin(n\pi))=1,$$




thus the limit points are: $$\liminf=0,1,2=\limsup$$

  • $\begingroup$ Wow, thanks for the fast answer. I already upvoted it, will accept it later to give others a chance as well :) especially for the liminf/limsup part. $\endgroup$ – John W. Dec 30 '16 at 14:28
  • $\begingroup$ @JohnW. are you still waiting. happy new year. $\endgroup$ – hamam_Abdallah Dec 31 '16 at 20:23

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