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Let $c_n=\left[\left|(-1)^n\left(\frac{n+1}{n}\right)\right|\right].$

I am stuck in finding $\inf(c_n)$ and $\sup(c_n).$

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    $\begingroup$ Are you using $[x]$ for the floor (greatest integer) function, nowadays usually written $\lfloor x\rfloor$? $\endgroup$ – Brian M. Scott Jun 5 '16 at 20:37
  • $\begingroup$ yes i am using for the floor (greatest integer). $\endgroup$ – Raio Jun 5 '16 at 20:40
  • $\begingroup$ Isn't that sequence constant for $n\ge 2$? Like, $$2,1,1,1,1,1,1,\ldots$$ (And what's the point of putting $(-1)^n$ inside an absoulte value?) $\endgroup$ – user228113 Jun 5 '16 at 21:26
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HINT: First note that

$$(-1)^n\left(\frac{n+1}n\right)=(-1)^n\left(1+\frac1n\right)\;.$$

Now $1<1+\frac1n<2$ for $n\ge 2$, so $-2<-\left(1+\frac1n\right)<-1$ for $n\ge 2$, and we have

$$c_n=\left\lfloor 1+\frac1n\right\rfloor=1$$

when $n\ge 2$ is even and

$$c_n=\left\lfloor-\left(1+\frac1n\right)\right\rfloor=-2$$

when $n\ge 2$ is odd. If you combine these two facts with a correct calculation of $c_1$, you should have little trouble determinine $\sup_nc_n$ and $\inf_nc_n$.

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If $\mid A \mid$ means the absolute of $A$ in this case, then $\inf(c_n)=1$ (just consider $n\to\infty$). And $\sup(c_n)=\max(c_n)=2$ for $n=1$.

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