1. Convergence of Sequence

Prove that if the sequence a(n+k) converges to l then the sequence a(n) also converges to l.

My proof basically looks like this:
there exists N s.t. n+k>n>N.
Assume the sequence a(n+k) converges to l and that the sequence a(n) diverges. By the definition: there exists L>0 s.t. for every K>0, there exists N that is an integer s.t n which is also an integer n>N -> |a(n)-L|<K

This means that if the sequence a(n) converges, then the sequence a(n+k) also converges because n+k>n>N. But if a(n) diverges, by the definition of divergence (eg not(definition of convergence)) a(n+k) also diverges. And since a(n+k) only converges when a(n) converges, there is a contradiction with my initial assumption where a(n+k) converges and a(n) diverges.

The problem that im having is with my first line, there exists N s.t. n+k>n>N. Is this a valid statement? I keep getting the feeling that there is something wrong with it.

2. Hi

The problem that im having is with my first line, there exists N s.t. n+k>n>N. Is this a valid statement? I keep getting the feeling that there is something wrong with it.
It is valid (if $k,\,n>0$) but I don't understand why you need it. Is it to say that if $n>N$ then $n+k>N$ ?

Originally Posted by ah-bee
Prove that if the sequence a(n+k) converges to l then the sequence a(n) also converges to l.

My proof basically looks like this:
there exists N s.t. n+k>n>N.
Assume the sequence a(n+k) converges to l and that the sequence a(n) diverges. By the definition: there exists L>0 s.t. for every K>0, there exists N that is an integer s.t n which is also an integer n>N -> |a(n+k)-L|<K

This means that if the sequence a(n) converges, then the sequence a(n+k) also converges because n+k>n>N.
Why not using this method to prove directly that if $(a_{n+k})$ converges then $(a_n)$ converges too ? I think it would make your proof clearer :

Let's assume that $(a_{n+k})$ converges, that is to say that $\forall \varepsilon,\,\exists N\,|\,n>N \implies |a_{n+k}-L|<\varepsilon$ hence by taking $N_0=?$ we get that $\forall \varepsilon,\,\exists N_0\,|\,n>N_0 \implies |a_{n}-L|<\varepsilon$ hence $(a_n)$ converges.

It's exactly the idea you used but the proof is both shorter and clearer.

Hope that helps.