Question Number 209514 by depressiveshrek last updated on 12/Jul/24
$$\mathrm{If}\:\Sigma\:{a}_{{n}} \:\mathrm{is}\:\mathrm{absolutely}\:\mathrm{convergent},\:\mathrm{prove}\:\mathrm{that} \\ $$$$\Sigma\:\frac{{a}_{{n}} }{{n}}\:\mathrm{is}\:\mathrm{also}\:\mathrm{absolutely}\:\mathrm{convergent}. \\ $$
Answered by MM42 last updated on 13/Jul/24
$$\frac{\mid{a}_{{n}} \mid}{{n}}\:<\mid{a}_{{n}} \mid\Rightarrow\Sigma\mid\frac{{a}}{{n}}\mid<\Sigma\mid{a}_{{n}} \mid \\ $$$$\Sigma\mid{a}_{{n}} \mid\:{is}\:{convergent}\:\Rightarrow\:\Sigma\mid\frac{{a}_{{n}} }{{n}}\mid\:\:{is}\:{convergent} \\ $$
Commented by depressiveshrek last updated on 12/Jul/24
$$\mathrm{Not}\:\mathrm{true},\:\mathrm{what}\:\mathrm{if}\:{a}_{{n}} \:\mathrm{is}\:\mathrm{decreasing}\:\mathrm{or}\:\mathrm{negative}, \\ $$$$\mathrm{then}\:\mathrm{that}\:\mathrm{inequality}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{hold}. \\ $$