How to Prove 0.333... < x/y < 0.5 with $x$ and $y$ Defined as Fractions?

[anemone]

Let $y=\dfrac{1}{10^2}+\dfrac{1}{11^2}+\cdots+\dfrac{1}{19^2}$ and $x=\dfrac{1}{21^2}+\dfrac{1}{22^2}+\cdots+\dfrac{1}{40^2}$.

Prove that $\dfrac{1}{3}\lt \dfrac{x}{y}\lt \dfrac{1}{2}$.

 

[kaliprasad]

Let $y=\dfrac{1}{10^2}+\dfrac{1}{11^2}+\cdots+\dfrac{1}{19^2}$ and $x=\dfrac{1}{21^2}+\dfrac{1}{22^2}+\cdots+\dfrac{1}{40^2}$.

Prove that $\dfrac{1}{3}\lt \dfrac{x}{y}\lt \dfrac{1}{2}$.

let me prove the 2nd part that is

$\dfrac{x}{y}\lt \dfrac{1}{2}$
or $2x < y$

we have

Spoiler

$\dfrac{1}{n^2}=\dfrac{4}{(2n)^2} > \dfrac{2}{(2n+1) ^2} +\dfrac{2}{(2n+2) ^2}$
by taking n from 10 to 19 and adding we get the result

 

[Albert1]

Let $y=\dfrac{1}{10^2}+\dfrac{1}{11^2}+\cdots+\dfrac{1}{19^2}$ and $x=\dfrac{1}{21^2}+\dfrac{1}{22^2}+\cdots+\dfrac{1}{40^2}$.

Prove that $\dfrac{1}{3}\lt \dfrac{x}{y}\lt \dfrac{1}{2}$.

Spoiler

$\dfrac{20}{861}=\dfrac {1}{21}-\dfrac{1}{41}<x<\dfrac {1}{20}-\dfrac{1}{40}=\dfrac{1}{40}\\$
$\dfrac{40}{861}<2x<\dfrac {1}{20}\\$
$\therefore \dfrac {1}{2x}>20----(A)\\$
$3x>\dfrac{60}{861}---(B)\\$
$\dfrac{1}{20}=\dfrac {1}{10}-\dfrac{1}{20}<y<\dfrac {1}{9}-\dfrac{1}{19}=\dfrac{10}{171}\\$
$\therefore y>\dfrac{1}{20}----(C)\\$
$\dfrac {1}{y}>\dfrac{171}{10}---(D)$
$(C)\times(A)>1$, and $(B)\times (D)>1$
the proof of both sides is finished

 

[kaliprasad]

Spoiler

$\dfrac{20}{861}=\dfrac {1}{21}-\dfrac{1}{41}<x<\dfrac {1}{20}-\dfrac{1}{40}=\dfrac{1}{40}\\$

How ?

 

[Albert1]

How ?

Spoiler

$\sum (\dfrac {1}{n}-\dfrac {1}{n+1})=\sum \dfrac {1}{n(n+1)}<\sum \dfrac {1}{n^2}<\sum \dfrac {1}{(n-1)(n)}=\sum (\dfrac {1}{n-1}-\dfrac {1}{n})$

 

[anemone]

Thanks Albert for participating and your method is correct.

Thanks to kaliprasad as well for proving half of the problem.:)

 

[anemone]

My solution:

Spoiler

First notice that $22^2>21^2>4(10^2),\,24^2>23^2>4(11^2),\cdots,\,40^2>39^2>4(19^2)$ we get $\dfrac{1}{4(10^2)}+\dfrac{1}{4(10^2)}+\dfrac{1}{4(11^2)}+\dfrac{1}{4(11^2)}+\cdots+\dfrac{1}{4(19^2)}+\dfrac{1}{4(19^2)}>\dfrac{1}{21^2}+\dfrac{1}{22^2}+\cdots+\dfrac{1}{40^2}$, i.e. $\dfrac{1}{2}\left(y\right)>x$ or simply $y>2x$.

On the other hand, we have $6(10^2)>22^2>21^2,\,6(11^2)>24^2>23^2,\cdots,\,6(19^2)>40^2>39^2$ we get $\dfrac{1}{6(10^2)}+\dfrac{1}{6(10^2)}+\dfrac{1}{6(11^2)}+\dfrac{1}{6(11^2)}+\cdots+\dfrac{1}{6(19^2)}+\dfrac{1}{6(19^2)}<\dfrac{1}{21^2}+\dfrac{1}{22^2}+\cdots+\dfrac{1}{40^2}$, i.e. $\dfrac{1}{3}\left(y\right)<x$ or simply $y<3x$.

Combining both results we get $\dfrac{1}{3}\lt \dfrac{x}{y}\lt \dfrac{1}{2}$ and we're hence done.