Inequality Proof: $\dfrac{2015}{(k+1)(k+4030)}<\dfrac{1}{k+1}$

[anemone]

Suppose $k>0$. Show that $\dfrac{2015}{(k+1)(k+4030)}<\dfrac{1}{k+1}-\dfrac{1}{k+2}+\dfrac{1}{k+3}-\dfrac{1}{k+4}+\cdots+\dfrac{1}{k+4029}-\dfrac{1}{k+4030}$.

 

[Opalg]

Suppose $k>0$. Show that $\dfrac{2015}{(k+1)(k+4030)}<\dfrac{1}{k+1}-\dfrac{1}{k+2}+\dfrac{1}{k+3}-\dfrac{1}{k+4}+\cdots+\dfrac{1}{k+4029}-\dfrac{1}{k+4030}$.

[sp]This is a special case ($n=2015$) of the inequality $$\frac n{(k+1)(k+2n)} < \frac{1}{k+1}-\dfrac{1}{k+2}+\dfrac{1}{k+3}-\dfrac{1}{k+4}+\ldots+\dfrac{1}{k+2n-1}-\dfrac{1}{k+2n},$$ which can be proved by induction for $n\geqslant2$.

In fact, the result almost holds for $n=1$ too, except that in this case the inequality becomes an equality, because $\dfrac1{(k+1)(k+2)} = \dfrac1{k+1} - \dfrac1{k+2}.$ So we can take this as the base case, provided that the proof of the inductive step involves a strict inequality. That will establish that the strict inequality occurs when $n=2$.

So suppose that the inequality holds for some value of $n$. Then $$\frac n{(k+1)(k+2n)} + \dfrac1{k+2n+1} - \dfrac1{k+2n+2} < \frac{1}{k+1}-\dfrac{1}{k+2}+\dfrac{1}{k+3}-\dfrac{1}{k+4}+\ldots+\dfrac{1}{k+2n+1}-\dfrac{1}{k+2n+2}.$$ Since $\dfrac1{k+2n+1} - \dfrac1{k+2n+2} = \dfrac1{(k+2n+1)(k+2n+2)},$ the left side of that inequality is equal to $\dfrac n{(k+1)(k+2n)} + \dfrac1{(k+2n+1)(k+2n+2)}.$

To establish the inductive step, we therefore need to show that $$\dfrac{n+1}{(k+1)(k+2n+2)} < \frac n{(k+1)(k+2n)} + \dfrac1{(k+2n+1)(k+2n+2)}.$$ Multiplying out the fractions, this is equivalent to showing that $$(n+1)(k+2n)(k+2n+1) < n(k+2n+1)(k+2n+2) + (k+1)(k+2n).$$ Now write both sides as polynomials in $k$ to see that this is equivalent to showing that $$(n+1)k^2 + (4n^2 + 5n + 1)k + 2n < (n+1)k^2 + (4n^2 + 5n + 1)k + 4n.$$ But since $4n>2n$, that last inequality obviously holds. That (somewhat inelegantly) establishes the inductive step and proves the result.[/sp]

 

[anemone]

Thanks for participating in this challenge, Opalg!

The solution that I have also prove the inequality with the induction method!(Smile)