Finally, the general case can be reduced to the case of n = 1 as follows : given a bounded sequence in \ mathbb { R } ^ n, the sequence of first coordinates is a bounded real sequence, hence has a convergent subsequence.
22.
In fact, the best solution is the interpolation result quoted by JackSchmidt, that is, as I understand, " a bounded sequence with convergent second Cesaro mean has the first Cesaro mean convergent too " .-- talk ) 23 : 26, 7 March 2009 ( UTC)
23.
The sequence 1 / n converges to zero and so is bounded . sin ( n ) / n is the product of two bounded sequences, so it's bounded . ( In fact, you can prove that it converges to zero . ) talk ) 14 : 31, 19 October 2014 ( UTC)
24.
For the first part of this, I tried using the fact that every coordinate is bounded and thus has a convergent subsequence ( is a bounded sequence ), so find a convergent subsequence in the first coordinate, then find a convergent subsequence of that sequence in the second coordinate, and then a convergent subsequence of that sequence in the third coordinate, and so on.
25.
Take a positive linear functional on the bounded sequences that is 0 if the sequence has only a finite number of nonzero elements and takes value 1 on the sequence 1, 1, 1, . . . ., so the functional in some sense gives an " average value " of any bounded sequence . ( Such a functional cannot be constructed explicitly, but exists by the Hahn Banach theorem . ) Then the content of a set of positive integers is the average value of the sequence that is 1 on this set and 0 elsewhere.
26.
Take a positive linear functional on the bounded sequences that is 0 if the sequence has only a finite number of nonzero elements and takes value 1 on the sequence 1, 1, 1, . . . ., so the functional in some sense gives an " average value " of any bounded sequence . ( Such a functional cannot be constructed explicitly, but exists by the Hahn Banach theorem . ) Then the content of a set of positive integers is the average value of the sequence that is 1 on this set and 0 elsewhere.
