Workshop Solutions 2.

1. Fix $\epsilon >0$. Observe that$\frac{1}{{\log }_2(n)}\le \epsilon$ if and only if ${\log }_2(n)\ge \frac{1}{\epsilon }.$ This is equivalent to say, that $n\ge 2^{\frac{1}{\epsilon }}.$ Hence, if $N=2^{\frac{1}{\epsilon }}$ then $|a_n|\le \epsilon$, whenever $n\ge N$.

2. Let $a_{2n-1}=1,a_{2n}=2.$ Also, let $b_{2n-1}=2,b_{2n}=1.$

3. [1] Clearly if a certain statement holds for all possible positive $\epsilon$, then there exists an $\epsilon >0$ for which the statement holds.

   [2] Let ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ be a Dauchy-sequence and $\epsilon >0$ and $N>0$ be as above. Then, if $m\ge N$: $|a_m|\le |a_N|+\epsilon$. On the other hand, let $K=\max \{|a_1|,|a_2|,\mathrm{\dots },|a_N|\}$. Then for any $i\ge 1$, $|a_i|\le K+\epsilon$.

   [3] Let the sequence ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ be bounded by $M>0$. Let $\epsilon =2M$, $N=1$. Clearly, if $n,m\ge N$ then by the triangle inequality, $|a_n-a_m|\le 2M=\epsilon$. That is, the sequence ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ is Dauchy.

4. [1] and [2] The answer is no. For any $n\ge 1$, let $x_n^n=n$, but let $x_n^k=0$ if $k\ne n$. Then for any given $k\ge 1$ $x_n^k\to 0$. On the other hand, $x_n^n\to \mathrm{\infty }$.

   [3] If , then, of course, as well. Hence by the Sandwich Lemma $x_k^k\to 0$.

   [4] Since for any $k\ge 1$ $x_n^k\to 0$, there exists some $i_k$ such that Therefore $x_{i_k}^k\to 0$.

   [5] Fix $\epsilon >0$. We need to show that there exists $N\ge 1$ such that if $n\ge N$: . Let $P>1$ such that if $n\ge P$ then $|a_n-a|\le \epsilon$. Set $N$ to be larger than $f^{-1}\{1,2,\mathrm{\dots },P\}$. So, if $n\ge N$, then $f(n)>P$. That is, if $n\ge N$, then $|a_{f(n)}-a|\le \epsilon .$