Product of convergent infinite series converges?

tarheelborn
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Homework Statement



Given two convergent infinite series such that \sum a_n -> L and \sum b_n -> M, determine if the product a_n*b_n converges to L*M.

Homework Equations





The Attempt at a Solution



If know that if a_n -> L this means that the sequence of partial sums of a_n = s_n converges to L. Similarly for the sequence of partial sums of b_n = t_n converges to M. I am not sure how to multiply these two sequences of partial sums.
 
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A_n = a_0 + a_1 + ... + a_n
B_n = b_0 + b_1 + ... + b_n

(A_n)(B_n) = (a_0 + a_1 + ... + a_n) (b_0 + b_1 + ... + b_n)
= sum of i=0 to n (inner sum of j = 0 to n) a_i b_j

Sorry I don't know how to use latex on this forum.
Does that help with multiplying the sequences?

so for example A_1 * B_1 = (a_0 + a_1) ( b_0 + b_1) = a_0b_0 + a_0b1 + a_1b_0 + a_1b_1
 
OK, that makes sense. Unfortunately, I still have no idea how to start this proof. I know that I have to do an epsilon proof that the limit is L*M, but it seems like I am going to need something more than that.
 
Maybe I'm wrong but I'll throw my idea at you=)

\sum_{k=0}^\infty a_k \mbox{\Rightarrow\ \forall\ \epsilon>0\ \exists\ N_1>0\ so\ \forall\ m>n>N_1, } |\sum_{k=n+1}^{m} a_k|< \epsilon \mbox{ and } |\sum_{k=0}^\infty a_k-L|<\epsilon

\sum_{k=0}^\infty b_k \mbox{\Rightarrow\ \forall\ \epsilon>0\ \exists\ N_2>0\ so\ \forall\ m>n>N_2, } |\sum_{k=n+1}^{m} b_k|< \epsilon \mbox{ and } |\sum_{k=0}^\infty b_k-M|<\epsilon

\mbox{So if I'm right, after } N> \max \{N_1, N_2\} \mbox{ good things should happen. =) }
 
But if I have a geometric series, say \sum (1/2^n) - > 2 and another geometric series, say \sum (3/5)^n -> 5/2. Now say I multiply these two series giving \sum ((3/10)^n). The product of the series converges, but it converges to 10/7 which is different from 2(5/2), so the conjecture cannot be proved. Thanks for your help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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