How do you prove this?

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1 Answer
George C.
Aug 7, 2018

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#.

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