If $\{b_{n}\}$ is obtained by deleting the first $k$ members of $\{a_{n}\}$, then $\{b_{n}\}$ is convergent iff $\{a_{n}\}$ is convergent.

Question

The question says:

Let $k \in N$. If the sequence $\{b_{n}\}$ is obtained by deleting the first $k$ members of the sequence $\{a_{n}\}$, then $\{b_{n}\}$ is convergent iff $\{a_{n}\}$ is convergent.

Could anyone give me a hint please?

Answer 1

The sequence $\{b_n\}$ is obtained from $\{a_n\}$ by shifting the indices $k$ steps. In other words, $b_{n} = a_{n+k}$

Suppose that $\lim a_n = L$, then

$$\forall\epsilon>0, \exists N \in \mathbb{N}: n\geq N\implies |a_n-L| < \epsilon$$

Hence, $n+k\geq N+k\implies |a_{n+k}-L| < \epsilon$. Set $N' = N + k$ and we have: $$\forall\epsilon>0, \exists N' \in \mathbb{N}:n\geq N'\implies |b_{n}-L| < \epsilon$$

Hence, $b_n \to L$. Conversely, suppose that $b_n \to L$

$$\forall\epsilon>0, \exists N \in \mathbb{N}: n\geq N\implies |b_n-L| < \epsilon$$

Since this holds for $N$, it holds for the larger value $N+k$ as well. Hence

$$\forall\epsilon>0, \exists N \in \mathbb{N}: n\geq N +k\implies |b_n-L| < \epsilon$$ Therefore, we have $n-k\geq N \implies |b_{n-k}-L| < \epsilon$. But $n-k \geq 1$ and $b_{n-k}=a_n$. Hence

$$\forall\epsilon>0, \exists N \in \mathbb{N}: n\geq N\implies |a_n-L| < \epsilon$$

Proving that $a_n \to L$.

Answer 2

Use the definition of convergence. Suppose $\lim\limits_{n\to\infty} a_n = a$. For all $\epsilon > 0$ there exists some $N$ such that $d(a_n,a)<\epsilon$ for all $n\geq N$. Now, modify $N$ accordingly.

Answer 3

$\{a_n\}_n$ converges to $A$ iff for every $r>0$ the set $\{n:a_n\not \in [-r+A, r+A]\}$ is finite.

If we delete the first $k$ terms of $\{a_n\}_n$ then we have a new series $\{b_n\}_n=\{a_{n+k}\}_n.$

For any $r>0$ we have $$\{n:b_n\not \in [-r+A,r+A]\} =$$ $$=\{n:a_{n+k}\not \in [-r+A,r+A]\}=$$ $$=\{n'>k :a_{n'}\not \in [-r+A,r+A]\}\subset$$ $$\subset \{n':a_{n'}\not\in [-r+A,r+A]\}.$$ This last set is finite.

Generalized: A sub-sequence of $\{a_n\}_n$ is $\{b_n\}=\{a_{f(n)}\}_n$ for some (any) strictly increasing $f:\Bbb N\to \Bbb N.$ If $\{a_n\}$ converges to $A$ then for any $r>0$ we have $$\{n:b_n\not \in [-r+A,r+A]\}=$$ $$=\{f(n'):b_{f(n')}\not\in [-r+A,r+A]\}=$$ $$=\{n'': \exists n'\;(\;f(n')=n''\land a_{n''}\not \in [-r+A,r+A]\;)\} \subset$$ $$\subset \{n'':a_{n''}\not \in [-r+A,r+A]\}$$ and this last set is finite.

In your Q, $\;f(n)=n+k.$