Convergent Series... Stumped
- Thread starter bearej50
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\(\displaystyle \text{DrMike inspired me . . .}\)
\(\displaystyle \text{Let: }\:\sum a_n \:=\:\sum \frac{(\text{-}1)^n}{\sqrt[3]{n}} \qquad \sum b_n \:=\:\sum\frac{(\text{-}1)^n}{\sqrt[3]{n^2}}\)
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\(\displaystyle \text{Both are alternating series whose }n^{th}\text{ terms }\to 0.\)
. . \(\displaystyle \text{Hence, both series converge.}\)
\(\displaystyle \text{But the series: }\:\sum a_nb_n \:=\:\sum\left(\frac{(\text{-}1)^n}{\sqrt[3]{n}}\right)\left(\frac{(\text{-}1)^n}{\sqrt[3]{n^2}}\right) \;=\;\sum\frac{1}{n}\)
. . \(\displaystyle \text{is the divergent Harmonic Series.}\)
\(\displaystyle \text{Let: }\:\sum a_n \:=\:\sum \frac{(\text{-}1)^n}{\sqrt[3]{n}} \qquad \sum b_n \:=\:\sum\frac{(\text{-}1)^n}{\sqrt[3]{n^2}}\)
\
\(\displaystyle \text{Both are alternating series whose }n^{th}\text{ terms }\to 0.\)
. . \(\displaystyle \text{Hence, both series converge.}\)
\(\displaystyle \text{But the series: }\:\sum a_nb_n \:=\:\sum\left(\frac{(\text{-}1)^n}{\sqrt[3]{n}}\right)\left(\frac{(\text{-}1)^n}{\sqrt[3]{n^2}}\right) \;=\;\sum\frac{1}{n}\)
. . \(\displaystyle \text{is the divergent Harmonic Series.}\)
