
#1
Sep2609, 09:27 PM

P: 200

1. The problem statement, all variables and given/known data
Assume that (an) is a bounded (but not necessarily convergent) sequence, and that the sequence (bn) converges to 0. Prove that the sequence (anbn) converges to zero. 2. Relevant equations 3. The attempt at a solution Assume that an is a bounded sequence and bn converges to 0. That means for all n in N, there exists a M >0 so that an<=M Since bn converges, that means that it must be bounded as well. Which means for all n in N there exists a P>0 so that bn<=P since an<=M and bn<=P that means for all n in N: anbn<= MP which is equivalent to anbn<=MP where MP>0 since M>0 and P>0. Hence (anbn) is bounded Since bn converges to 0 that means for e>0 there exists an N in N so that for n>=N bn0<e which is equivalent to e<bn<e This is where I get stuck. Do I just multiply the inequality by an? cause then I'd have e(an)<bnan<e(an) which would be equivalent to anbn<e2 if I let e2=e(an) which would mean that anbn converges to zero as well. But I don't know if I can multiply the sequence by it though.... Any help would be great! 



#2
Sep2609, 11:10 PM

HW Helper
P: 3,309

As b_{n} is convergent, I would say for any P>0, there exists N such that for all n>N then b_{n}0< b_{n} now what you need to show to prove a_{n}.b_{n} converges to zero, is that for any e>0 you can choose N, such that for all n>N you have a_{n}.b_{n}<e as you know a_{n}<=M for all n, then a_{n}.b_{n}<=M.b_{n} so now you just need to show you can choose N such that for all n>N b_{n}<=e/M and i think you're there 



#3
Sep2609, 11:15 PM

HW Helper
P: 3,309

updated above




#4
Sep2709, 11:38 AM

P: 200

converging analysis proof
thanks for the help :)



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