Is $X$ Larger Than $Y$ in POTW #257?

[anemone]

Here is this week's POTW:

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Which number, $X$ or $Y$, is larger?

$X=\dfrac{1}{2016}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2016}\right)$

$Y=\dfrac{1}{2017}\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{2017}\right)$

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[anemone]

Congratulations to the following members for their correct solution::)

1. Opalg
2. lfdahl
3. Theia

Solution from Opalg:

Spoiler

Multiply the first equation by $2016$, the second one by $2017$, and subtract: $$2017Y - 2016X = \dfrac1{2017}.$$ That shows that the point $(X,Y)$ lies on the line $y = \dfrac{2016}{2017}x + \dfrac1{2017^2}$ (the red line in the diagram, which is admittedly not at all drawn to scale). That line meets the line $y=x$ (the blue line) at the point $\Bigl(\dfrac1{2017},\dfrac1{2017}\Bigr)$. Since the red line has gradient just very slightly less than $1$, every point $(x,y)$ on it that lies above the point of intersection is below the blue line and will therefore satisfy $y<x$. But it is clear from the defining equation for $Y$ that $Y > \dfrac1{2017}$. So the point $(X,Y)$ does lie above the point of intersection, and therefore $Y<X$.
[TIKZ]\draw (-1,0) -- (4,0) ;
\draw (0,-1) -- ( 0,4) ;
\draw[thick, blue] (-1,-1) -- (4,4) ;
\draw[thick, red] (-1,-0.8) -- (4,3.7) ;
\fill (1,1) circle (2pt) ;
\draw (2,0.8) node {$\bigl(\frac1{2017},\frac1{2017}\bigr)$} ;
[/TIKZ]

Alternate solution from lfdahl:

Spoiler

\[\begin{align*}X-Y &= \left ( \frac{1}{2016}-\frac{1}{2017} \right )\left ( 1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2016} \right )-\frac{1}{2017^2} \\\\& =\frac{1}{2016\cdot 2017}\left ( 1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2016} \right )-\frac{1}{2017^2} \\\\& >\frac{1}{2017^2}\left ( 1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2016} \right )-\frac{1}{2017^2} \\\\& =\frac{1}{2017^2}\left ( \frac{1}{2}+\frac{1}{3}+..+\frac{1}{2016} \right ) \\\\&> 0 \end{align*}\]

Thus, $X$ is the larger of the two.