Question

How do you prove this?

Answer 1

See explanation...

Explanation:

(a) Suppose `lim_(n->oo) a_n = L`

Then by the definition of limit:

`AA epsilon > 0 EE N in NN : AA n > N, abs(a_n - L) < epsilon`

So given any `epsilon > 0`, choose `N_epsilon` such that `AA n > N_epsilon, abs(a_n - L) < epsilon`.

Then `AA n > N_epsilon / 2, abs(b_n - L) = abs(a_(2n) - L) < epsilon` (since `2n > N_epsilon`).

Therefore `b_n` satisfies the limit definition for `lim_(n->oo) b_n = L`.

(b) The converse to (a) would be:

Suppose `b_n` is a sequence of numbers with `lim_(n->oo) b_n = L` and `a_n` is a sequence of numbers such that `b_n = a_(2n)`. Then `lim_(n->oo) a_n = L`.

This is not true in general.

For example, consider:

`a_n = { (0 " if " n " is even"), (n " if " n " is odd") :}`

Then:

`b_n = a_(2n) = 0`

and hence:

`lim_(n->oo) b_n = 0`

but:

`lim_(n->oo) a_n`

does not converge to any limit.

A statement that would be true is that if `b_n = a_(2n)` and `a_n` does converge to a limit, then that limit is equal to `lim_(n->oo) b_n = L`.