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Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$

The book I am using asserts a non-trivial way of proving this inequality, but I cannot see why this cannot be proven by rearranging the statement equally as rigorously.

Let $$\frac{2k+2}{2k+3}=\frac{2k}{2k+1}, k\in\mathbb N$$ $$(2k+2)(2k+1)=2k(2k+3)$$ $$2k^2+3k+1=2k^2+3k$$ $$1=0$$ Contradiction, therefore no solution to where the two sides are equal

Let k=1 $$\frac{2+2}{2+3}= 0.8$$ $$\frac{2}{3}=0.66666$$ Since there are no points of intersection, and for k=1 LHS>RHS, and both sides are continuous on k>0, inequality must hold.

jamie
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    This proof is not right. You only prove there doesn't exist a $k$ such that the equality is reachable and that it is true for $k=1$. What about the others $k>1$? Continuity (which does not make much sense for a function whose domain is $\mathbb{N}$) is not an argument. – LHF Mar 09 '20 at 20:10
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    You're assuming that the two functions are monotonic. – Luke Collins Mar 09 '20 at 20:13
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    What happens if you replace $=$ with $>$ in your first argument? – Quang Hoang Mar 09 '20 at 20:14
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    As mentioned, your proof doesn't work properly as stated. However, $\frac{2k+2}{2k+3} = 1 - \frac{1}{2k+3}$ might be useful in a proof. – John Omielan Mar 09 '20 at 20:14
  • @Atticus Can you show me where exactly I go wrong in my intentions. I show that there is no intersection to show that LHS is either less than RHS, or greater than RHS for any value k. I then show it is true for k=1, to show that it is greater than RHS. Since I showed there were no intersections previously, LHS must be greater than RHS for any other k. – jamie Mar 09 '20 at 20:25
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    @jamie, the fractions are defined on $\mathbb{N}$ (they don't have to intersect to overpass each other) – LHF Mar 09 '20 at 20:32
  • @Atticus Is it not possible to ignore the domain and prove for a larger domain? Since what I showed holds for any real k, and N being a subset of R, it must also hold for N(allowing me to bring in continuity to show that they must intersect to overpass) – jamie Mar 09 '20 at 20:37
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    Yours works if you say something like "both sides extend to functions on $\mathbb{R}^{+}$"

    The other commentors are basically arguing that you can't use continuity when your domain of definition is discrete.

     – user113102 Mar 09 '20 at 20:40
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    @jamie, if you extend to $\mathbb{R}^+$ you would have to evaluate the limits to $\infty$ and see what happens there. Overall, studying a inequality defined on $\mathbb{N}$ with calculus arguments seems like a headache compared to Quang Hoang's suggestion. – LHF Mar 09 '20 at 20:42
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    I mean you also need to check your algebra there. – user113102 Mar 09 '20 at 20:42

3 Answers3

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Let's consider the following inequality verified for every k natural: $$\frac{2}{(2k+3)(2k+1)}\gt0$$ Now, with simple manipulations $$\frac{4k^2+6k+2-4k^2-6k}{(2k+3)(2k+1)}\gt0$$ $$\frac{(2k+2)(2k+1)-2k(2k+3)}{(2k+3)(2k+1)}\gt0$$ $$\frac{2k+2}{2k+3}-\frac{2k}{2k+1}\gt0$$ $$\frac{2k+2}{2k+3}\gt\frac{2k}{2k+1}$$

Augusto Santi
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$\frac{2x+2}{2x+3}>\frac{2x}{2x+1}, X\in\mathbb R^+$

$ f: x \rightarrow \frac{2x}{2x+1} = f: x \rightarrow 1- \frac{1}{2x+1}$ is easily demonstrable to be strictly increasing

$x > y \rightarrow 2x+1 > 2y+1 \rightarrow \frac{1}{2x+1} <\frac{1}{2y+1} \rightarrow 1- \frac{1}{2x+1} > 1-\frac{1}{2y+1}$ therefore $f(k+1) > f(k) $

i.e : $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N \subset \mathbb R^+$

Mohammed M. Zerrak
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Because $k\in N$, and the denominators are always positive, we have: $$(2k+2)(2k+1)>2k(2k+3)$$ So: $$4k^2+6k+2>4k^2+6k$$ Now, we can cancel out $4k^2$ and $6k$, obtaining:$$2>0$$ which is always true. The inequality holds.

Matteo
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