HW 2

Work is shown on page 449

#todo

Proof:

1. Assume $ϵ>0$
2. Then lets pick $N=\frac{3}{ϵ}-1$
3. For any $n>\frac{3}{ϵ}-1=N$
4. This implies that $n+1>\frac{3}{ϵ}$
5. Meaning, $\frac{3}{n+1}<ϵ$
6. Notice $\frac{3}{n+1}=\mid -\frac{3}{n+1}\mid =\mid -\frac{3}{n+1}\mid =\mid \frac{3n}{n+1}-3\mid <ϵ$
7. Thus For any $ϵ>0$ there exists an $N$ such that $n>N⟹\mid \frac{3n}{n+1}-3\mid <ϵ$. By the limit convergence theorem this holds.

Proof:

1. Assume $ϵ>0$
2. Then, lets pick $N=\max (1,\frac{1}{ϵ})$
3. Then, for $n>N$ we have $n>\frac{1}{ϵ}$
4. This implies that $ϵ>\frac{1}{n}$
5. Notice that $ϵ>\frac{1}{n}=\frac{n}{n^2}\ge \frac{n-1}{n^2+1}$
6. As $n>1$ can confidently say $\frac{n-1}{n^2+1}=\mid \frac{n-1}{n^2+1}\mid <ϵ$
7. Thus, we have shown for all $ϵ>0$ there exists an $N$ such that $n>N$ implies $\mid \frac{n-1}{n^2+1}-0\mid <ϵ$. By the convergence definition this sequence converges.

#todo

Proof:

1. Assume $s_n\to s$ and all $s_n\ge 0$
2. Then for all $ϵ>0$ we know there exists $N$ such that $n>N⟹|s_n-s|<ϵ$
3. For the sake of contradiction, assume $s<0$. Then we can set $ϵ=\frac{-s}{2}$
4. $|s_n-s|<\frac{-s}{2}⟹s_n-s<\frac{-s}{2}$.
5. Then, we have it so $s_n<s-\frac{s}{2}=\frac{s}{2}<0$. This is a contradiction as it implies $s_n<0$.