Give an example of two convergent series ?an, ?bn such that ?(anbn) diverges.
B bearej50 New member Joined Feb 16, 2009 Messages 21 May 10, 2009 #1 Give an example of two convergent series ?an, ?bn such that ?(anbn) diverges.
D DrMike Full Member Joined Mar 31, 2009 Messages 252 May 10, 2009 #2 bearej50 said: Give an example of two convergent series ?an, ?bn such that ?(anbn) diverges. Click to expand... Hmm.... make a[sub:27ywpii6]n[/sub:27ywpii6] and b[sub:27ywpii6]n[/sub:27ywpii6] alternating, maybe?
bearej50 said: Give an example of two convergent series ?an, ?bn such that ?(anbn) diverges. Click to expand... Hmm.... make a[sub:27ywpii6]n[/sub:27ywpii6] and b[sub:27ywpii6]n[/sub:27ywpii6] alternating, maybe?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 May 11, 2009 #3 DrMike inspired me . . .\displaystyle \text{DrMike inspired me . . .}DrMike inspired me . . . Let: ∑an = ∑(-1)nn3∑bn = ∑(-1)nn23\displaystyle \text{Let: }\:\sum a_n \:=\:\sum \frac{(\text{-}1)^n}{\sqrt[3]{n}} \qquad \sum b_n \:=\:\sum\frac{(\text{-}1)^n}{\sqrt[3]{n^2}}Let: ∑an=∑3n(-1)n∑bn=∑3n2(-1)n \ Both are alternating series whose nth terms →0.\displaystyle \text{Both are alternating series whose }n^{th}\text{ terms }\to 0.Both are alternating series whose nth terms →0. . . Hence, both series converge.\displaystyle \text{Hence, both series converge.}Hence, both series converge. But the series: ∑anbn = ∑((-1)nn3)((-1)nn23) = ∑1n\displaystyle \text{But the series: }\:\sum a_nb_n \:=\:\sum\left(\frac{(\text{-}1)^n}{\sqrt[3]{n}}\right)\left(\frac{(\text{-}1)^n}{\sqrt[3]{n^2}}\right) \;=\;\sum\frac{1}{n}But the series: ∑anbn=∑(3n(-1)n)(3n2(-1)n)=∑n1 . . is the divergent Harmonic Series.\displaystyle \text{is the divergent Harmonic Series.}is the divergent Harmonic Series.
DrMike inspired me . . .\displaystyle \text{DrMike inspired me . . .}DrMike inspired me . . . Let: ∑an = ∑(-1)nn3∑bn = ∑(-1)nn23\displaystyle \text{Let: }\:\sum a_n \:=\:\sum \frac{(\text{-}1)^n}{\sqrt[3]{n}} \qquad \sum b_n \:=\:\sum\frac{(\text{-}1)^n}{\sqrt[3]{n^2}}Let: ∑an=∑3n(-1)n∑bn=∑3n2(-1)n \ Both are alternating series whose nth terms →0.\displaystyle \text{Both are alternating series whose }n^{th}\text{ terms }\to 0.Both are alternating series whose nth terms →0. . . Hence, both series converge.\displaystyle \text{Hence, both series converge.}Hence, both series converge. But the series: ∑anbn = ∑((-1)nn3)((-1)nn23) = ∑1n\displaystyle \text{But the series: }\:\sum a_nb_n \:=\:\sum\left(\frac{(\text{-}1)^n}{\sqrt[3]{n}}\right)\left(\frac{(\text{-}1)^n}{\sqrt[3]{n^2}}\right) \;=\;\sum\frac{1}{n}But the series: ∑anbn=∑(3n(-1)n)(3n2(-1)n)=∑n1 . . is the divergent Harmonic Series.\displaystyle \text{is the divergent Harmonic Series.}is the divergent Harmonic Series.
D DrMike Full Member Joined Mar 31, 2009 Messages 252 May 11, 2009 #4 Ack! Now you've given him the answer!
