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The question is to prove that $\dfrac{1+ny}{1-y^{n+1}}-\dfrac{1}{1-y}$ is a decreasing function in $y$ for $y>1$, where $n$ is a positive integer.

My first thought is to take the derivative and show the sign is negative. But it seems the derivative is a bit complicate to easily draw the conclusion. Any other thoughts/hints/comments? Thanks a lot.

Amzoti
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ipmser
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    Consider multiplying the second fraction by $\sum_{i=0}^n y^i$ in numerator and denominator, then adding the fractions directly... – abiessu Jan 06 '14 at 19:10
  • Thanks for the hint. I can see this is very helpful in showing $\frac{1+ny}{1-y^{n+1}}-\frac{1}{1-y}$ is positive, but not sure how the monotonicity can be addressed from this trick. – ipmser Jan 06 '14 at 19:35
  • Haven't solved this problem, but having spent a little time on it with my own tricks, I rather like it. – abnry Jan 07 '14 at 21:51

1 Answers1

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Beginning with

$$f(y)=\dfrac{1+ny}{1-y^{n+1}}-\dfrac{1}{1-y}=\dfrac{1+ny}{1-y^{n+1}}-\dfrac{\sum_{i=0}^ny^i}{1-y^{n+1}}\\ =\dfrac{(n-1)y-\sum_{i=2}^ny^i}{1-y^{n+1}}=\frac{\sum_{i=2}^ny^i-(n-1)y}{y^{n+1}-1}$$

the derivative is

$$f'(y)=\frac {(y^{n+1}-1)(\sum_{i=2}^niy^{i-1}-n+1)-(n+1)y^n(\sum_{i=2}^ny^i-(n-1)y)}{(n+1)^2y^{2n}}$$

EDIT: This is too messy to work with properly. With a restart from the beginning, we have

$$f'(y)=\frac {n(1-y^{n+1})-(1+ny)(n+1)y^n}{(n+1)^2y^{2n}}-\frac 1{(1-y)^2}$$

$$=\frac{n-ny^{n+1}-ny^n-y^n-n^2y^{n+1}-ny^{n+1}}{(n+1)^2y^{2n}}-\frac 1{(1-y)^2}$$

which is all-negative terms except for the initial $n$, and given that $y\gt 1$ we have $ny^n\gt n$ which means that the first fraction is always negative for $n\ge 2,y\gt 1$, and the second fraction is a negative of a square, which is always non-positive. Therefore, the derivative of $f(y)$ is negative for all $n\ge 2$ and for all $y\gt 1$.

abiessu
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  • That cannot be correct. If you substitute $n = 2$ into the original function, you get $y/(1+y+y^2)$, and its derivative is $(1-y^2)/(1+y+y^2)^2$, which has a critical point at $y = 1$, and for $y > 1$, $1-y^2 < 0$. – heropup Jan 06 '14 at 20:47
  • @heropup: I have removed a lot of the messy work and found the required result. How does it look now? – abiessu Jan 07 '14 at 20:52