Recent content by Checko

I understand the multiplying a series with a sequence now. Back to answering the problem We know b_n is bounded IFF there exist M s.t. b_n ≤ M If ∑a_n is absolutely convergent, then ∑a_n is convergent ∑a_n*b_n ≤ ∑a_n*M ∑a_n*b_n ≤ M∑a_n Since ∑a_n convergers then M∑a_n converges...

Let Sum a_n converge absolulty Let b_n be bounded w.w.t.s. (we wish to show) A) (Sum a_n) *(b_n) converges My case was a_n =1/n^2 b_n =(-1)^n B) As (sum (a_n*b_n)) it converges As b_n(sum a_n) it diverges due to the alternating -1 being outside the sum When given A (w/ or w/o...

Is the following true? (any sequence)(any series) make the sequence part of the series?

I found a case were the summation of "a sub n" is absolute convergent times a "b sub n" that is bounded does not converge (Sumation 1/n^2) (-1)^n This product is bounded but does not converge. Do I just show this example as proof that original question is false?

Homework Statement Let Sumation "a sub n" be an absolutely convergent series, and "b sub n" a bounded sequence. Prove that sumation "a sub n"*"b sub n" is convergent. (sorry fist time on this site and can't use the notation.) Homework Equations Theorem A: that states sumation "a sub n"...