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

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The discussion centers on comparing two mathematical expressions, $X$ and $Y$, defined using harmonic sums. $X$ is calculated as $\dfrac{1}{2016}$ times the sum of the first 2016 harmonic numbers, while $Y$ is $\dfrac{1}{2017}$ times the sum of the first 2017 harmonic numbers. Participants engage in solving the problem and share their solutions, with Opalg providing the main solution and lfdahl offering an alternate approach. The thread highlights the importance of understanding harmonic series in determining the relationship between $X$ and $Y$. Overall, the conversation emphasizes mathematical reasoning and problem-solving techniques.
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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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Congratulations to the following members for their correct solution::)

1. Opalg
2. lfdahl
3. Theia

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

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