What is the relationship between A and B?

Question

We have $0<b≤ a$, and:

$$\underbrace{\dfrac{1+⋯+a^7+a^8}{1+⋯+a^8+a^9}}_{A} \quad \text{and} \quad \underbrace{\dfrac{1+⋯+b^7+b^8}{1+⋯+b^8+b^9}}_{B}$$

Source: Lumbreras Editors

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It was my strategy:

$1 ≤ \dfrac{1+⋯+a^8}{1+⋯+b^8}$ ya que $a^p ≥ b^p$, $∀\, p ≥ 0 $ (monotonous function).

We also have to, $ \dfrac{a}{b} ≤ 1 $

Answer 1

$$1-A=1-\frac{1+a+...+a^8}{1+a+...+a^9}=\frac{a^9}{1+a+...+a^9}=$$ $$=\frac{1}{\frac{1}{a^9}+\frac{1}{a^8}+...+1}\geq\frac{1}{\frac{1}{b^9}+\frac{1}{b^8}+...+1}=1-B,$$ which gives $$A\leq B.$$

Answer 2

Let $$ f(x)=\dfrac{1+⋯+x^7+x^8}{1+⋯+x^8+x^9}. $$ Then $$ f'(x)=-\frac{x^8(x^8+2x^7+3x^6+4x^5+5x^4+6x^3+7x^2+8x+9)}{(1+⋯+x^8+x^9)^2}\le0 $$and hence $f(x)$ is decreasing. So if $a\ge b$, then $f(a)\le f(b)$.

Answer 3

Step 1: Take the inverses, so we want to compare

$$ \frac { 1+a+\ldots a^9} { 1+a+\ldots a^8 } \text{ vs } \frac{1+b+\ldots b^9 }{1+b+\ldots b^8 } . $$

Step 2: Subtract 1 from both sides, we want to compare

$$ \frac { a^9} { 1+a+\ldots a^8 } \text{ vs } \frac{ b^9 }{1+b+\ldots b^8 } .$$

Step 3: Comparing these quantities is obvious by cross multiplying (denominators are positive), and using the fact that $ a^9 b^k > a^k b^9$ for $ k < 9$, so

$$ \frac { a^9} { 1+a+\ldots a^8 } > \frac{ b^9 }{1+b+\ldots b^8 } .$$

(Go back and fill in the gaps yourself.) Hence,

$$ \frac { 1+a+\ldots a^8} { 1+a+\ldots a^9 } < \frac{1+b+\ldots b^8 }{1+b+\ldots b^0 } . $$

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Notes

• Please show your work. When comparing fractions. a good tactic to use is cross multiplying. Nothing harder than that was used here.
• If we cross multiplied directly at the start, then subtracting 1 in step 2 is equivalent to removing common terms of $(1+a+\ldots+ a^8)(1+b+\ldots + b^8 )$.
• This approach generalizes to show that $ \frac{ a^n - 1 } { a^m -1 } > \frac{ b^n -1 } { b^m - 1 } $ if $(a-b)(n-m) > 0$.