Is Sigma(y(n)) Absolutely Convergent if y(n) = O(x(n))?

penguin007
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Hi,

assume sigma(x(n)) is an absolutly convergent series and that y(n)=O(x(n)), then can we conclude that sigma(y(n)) is absolutly convergent?


thanks
 
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Yes. That is pretty close to being the definition of "O".
 
thanks again
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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