True or false? $\limsup_{n\rightarrow\infty}\left((n \mod 4)+\frac{1}{n+1} \right )=\frac{9}{2}$

Question

Is the statement true or false?

$$\limsup_{n\rightarrow\infty}\left((n \bmod 4)+\frac{1}{n+1} \right )=\frac{9}{2}$$

No proofs needed, just an answer but I'd like to know the reason, too.

So lim sup means the greatest cluster point of the given sequence, right?

I know that $\frac{1}{n+1}$ will go to $0$ for $n\rightarrow\infty$ (corrected mistake).

Because I don't see why $n \bmod 4$ added with $0$ should equal $\frac{9}{2}$, I would say that the statement is false.

What do you think about this?

Answer 1

Since $n\bmod 4\le 3$, we have that, for $n>0$, $$ (n\bmod 4)+\frac{1}{n+1}<4<\frac{9}{2} $$

Answer 2

??? I know that $\frac1{n+1}$ will go to 1 for $n→∞$.

Ermm I don't know that...

The statement is false anyway; $\frac92$ is bigger than a much simpler upper bound of $4$.

Answer 3

False, the answer is $3$.

Proof.

$n$ mod $4$ has only four possibilities: $0,1,2,3$ and $\frac{1}{n+1}$ goes to $0$.

Answer 4

False.

The sequence $$a_n=n \!\!\!\!\!\mod 4+\frac{1}{n+1}$$ does NOT converge, and it possesses 4 subsequential limits:

$$ a_{4n}=\frac{1}{n+1}\to 0,\quad a_{4n+1}=1+\frac{1}{n+1}\to 1,\\ a_{4n+2}=2+\frac{1}{n+1}\to 2,\quad a_{4n+3}=3+\frac{1}{n+1}\to 3. $$