Is this a proof rigorous enough? (Definition of the limit)

Question

Prove that if a sequence $c_n$ converges to a limit L, then so does the same sequence with even $n$:

If $\displaystyle{\lim_{n \to \infty}}$ $c_n = L$ then $\forall \epsilon >0, \exists N>0$ such that $n>N \implies$ $|c_n-L|<\epsilon$

If this is true, then it is true for all integer choices of n. So it is true for even $n$ : $n=2k$, $k \in \textbf{N} $ .

$\therefore$ $2k >N \implies |c_{2k}-L|<\epsilon$

$\therefore$ for all $\epsilon>0$,$\exists M$ such that:

$k>M \implies |c_{2k}-L|<\epsilon$

$\therefore \displaystyle{\lim_{n \to \infty}}$ $c_{2n} = L$

Answer 1

You did not explain why such $M$ exists. It's not hard, though. You may take any natural number greater than or equal to $\frac N2$.

Answer 2

Given that

• $\displaystyle{\lim_{n \to \infty}}$ $c_n = L$ then $\forall \epsilon >0, \exists N$ such that $n>N \implies$ $|c_n-L|<\epsilon$

let consider $\displaystyle{\lim_{k \to \infty}}$ $c_{2k}$ then

• $\forall \epsilon >0, \exists N$ such that $2k>N, k>\frac{N}2 \implies$ $|c_{2k}-L|<\epsilon$

thus

$$\displaystyle{\lim_{k \to \infty}} c_{2k}=L$$