How would you show this summation is greater than 24 without induction?

[MarkFL]

On another site, a user asked for help showing:

$\displaystyle \sum_{k=0}^{2499} \frac{1}{\sqrt{4k+1}+\sqrt{4k+3}}>24$

The first respondent asked if the OP was familiar with mathematical induction. The reply was that induction was the topic of the next chapter in her course.

Another suggested rationalizing the denominator to write:

$\displaystyle \frac{1}{2}\sum_{k=0}^{2499} \left(\sqrt{4k+3}-\sqrt{4k+1} \right)>24$

and then wished the OP good luck. She then asked for further help. She also asked for a proof by induction, which I provided as follows:

If I were going to use induction, I would state the hypothesis:

$\displaystyle \frac{1}{2}\sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{n+1}-1$

or equivalently:

$\displaystyle \sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{n+1}-2$ where $\displaystyle n\in\mathbb{N}_0$

base case $\displaystyle P_0$:

$\displaystyle \sum_{k=0}^0\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{0+1}-2$

$\displaystyle \sqrt{3}-\sqrt{1}>1-2$ true.

Consider:

$\displaystyle 6n+9=2\left(3n+\frac{9}{2} \right)$

$\displaystyle 6n+9=2\left(2(2n+3)-\frac{2n+3}{2} \right)$

$\displaystyle 6n+9=2\left(\sqrt{4(2n+3)^2}-\sqrt{\frac{(2n+3)^2}{4}} \right)$

$\displaystyle 6n+9>2\left(\sqrt{4(2n+3)^2-1}-\sqrt{\frac{(2n+3)^2+4n+3}{4}} \right)$

$\displaystyle 6n+9>2\left(\sqrt{(4n+5)(4n+7)}-\sqrt{(n+1)(n+3)} \right)$

$\displaystyle 8n+12-2\sqrt{(4n+5)(4n+7)}>2n+3-2\sqrt{(n+1)(n+3)}$

$\displaystyle (4n+7)-2\sqrt{(4n+5)(4n+7)}+(4n+5)>(n+2)-2\sqrt{(n+1)(n+3)}+(n+1)$

$\displaystyle (\sqrt{4n+7}-\sqrt{4n+5})^2>(\sqrt{n+2}-\sqrt{n+1})^2$

$\displaystyle \sqrt{4n+7}-\sqrt{4n+5}>\sqrt{n+2}-\sqrt{n+1}$

$\displaystyle \sqrt{4(n+1)+3}-\sqrt{4(n+1)+1}>\sqrt{(n+1)+1}-\sqrt{n+1}$

Now, adding this to the hypothesis, we have:

$\displaystyle \sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)+\sqrt{4(n+1)+3}-\sqrt{4(n+1)+1}>\sqrt{n+1}-2+\sqrt{(n+1)+1}-\sqrt{n+1}$

$\displaystyle \sum_{k=0}^{n+1}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{(n+1)+1}-2$

$\displaystyle \frac{1}{2}\sum_{k=0}^{n+1}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{(n+1)+1}-1$

We have derived $\displaystyle P_{n+1}$ from $\displaystyle P_n$, thereby completing the proof by induction, and we may now state:

$\displaystyle \frac{1}{2}\sum_{k=0}^{2499}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{2500}-1=24$

Another person replied with a technique using integration, which I am sure will be of little use to the OP.

I am just curious if there is a way to demonstrate the inequality is true by purely algebraic means. This is just for my own curiosity, and I will not post anyone's work there.

 

[Opalg]

On another site, a user asked for help showing:

$\displaystyle \sum_{k=0}^{2499} \frac{1}{\sqrt{4k+1}+\sqrt{4k+3}}>24$

The first respondent asked if the OP was familiar with mathematical induction. The reply was that induction was the topic of the next chapter in her course.

Another suggested rationalizing the denominator to write:

$\displaystyle \frac{1}{2}\sum_{k=0}^{2499} \left(\sqrt{4k+3}-\sqrt{4k+1} \right)>24$

My starting point is the inequality $$\sqrt x > \tfrac12\bigl(\sqrt{x+\alpha}+\sqrt{x-\alpha}\bigr)\qquad (0<\alpha < x)\qquad(*).$$ Geometrically, this is an obvious consequence of the fact that the graph of the square root function is convex downwards, as in this picture:

[graph]jcs7w6tdh4[/graph]​

To verify (*) algebraically, square both sides so that it becomes $x > \frac14\bigl(2x +2\sqrt{x^2-\alpha^2}\bigr)$. This simplifies to $x>\sqrt{x^2-\alpha^2}$ which is obviously true.

Next, check that (*) can be written in the equivalent form $\sqrt x - \sqrt{x-\alpha} >\tfrac12\bigl(\sqrt{x+\alpha}-\sqrt {x-\alpha}\bigr).$ Now put $x=k+\frac34$ and $\alpha=\frac12$, to get $\sqrt{k+\frac34} - \sqrt{k+\frac14} >\tfrac12\bigl(\sqrt{k+\frac54}-\sqrt {k+\frac14}\bigr).$

Therefore $$\sum_{k=0}^{2499} \frac{1}{\sqrt{4k+1}+\sqrt{4k+3}} = \frac{1}{2}\sum_{k=0}^{2499} \left(\sqrt{4k+3}-\sqrt{4k+1} \right) = \sum_{k=0}^{2499} \left(\sqrt{k+\tfrac34}-\sqrt{k+\tfrac14} \right) > \frac{1}{2}\sum_{k=0}^{2499} \left(\sqrt{k+\tfrac54}-\sqrt{k+\tfrac14} \right).$$ That is a telescoping sum which collapses to $\frac12\Bigl(\sqrt{2500\tfrac14} - \sqrt{\tfrac14}\Bigr) > \frac12\bigl(50 - \tfrac12\bigr) = 24.75.$

 

[Poly1]

If I were going to use induction, I would state the hypothesis:

$\displaystyle \frac{1}{2}\sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{n+1}-1$

How did you get the RHS?

 

[MarkFL]

I noticed the partial sums of the LHS followed the curve $\displaystyle y=\frac{1}{2}\sqrt{x}$ very closely, so I wrote 24 in that form.

 

[CellCoree]

Yes, there is a way to demonstrate the inequality is true by purely algebraic means. We can use the AM-GM inequality to show that each term in the summation is greater than or equal to 1, and then use the fact that there are 2500 terms in the summation to show that the total sum is greater than 2500.

First, we can rewrite the summation as:

$\displaystyle \sum_{k=0}^{2499} \frac{1}{\sqrt{4k+1}+\sqrt{4k+3}} = \frac{1}{2} \sum_{k=0}^{2499} \left(\frac{1}{\sqrt{4k+1}} - \frac{1}{\sqrt{4k+3}}\right)$

Notice that each term in the summation can be written as:

$\displaystyle \frac{1}{\sqrt{4k+1}} - \frac{1}{\sqrt{4k+3}} = \frac{\sqrt{4k+3} - \sqrt{4k+1}}{\sqrt{(4k+1)(4k+3)}}$

Now, we can use the AM-GM inequality to show that the numerator is greater than or equal to 1. This is because:

$\displaystyle \sqrt{4k+3} - \sqrt{4k+1} = \frac{4k+3 - (4k+1)}{\sqrt{4k+3} + \sqrt{4k+1}} = \frac{2}{\sqrt{4k+3} + \sqrt{4k+1}}$

And by the AM-GM inequality, we have:

$\displaystyle \frac{2}{\sqrt{4k+3} + \sqrt{4k+1}} \geq \frac{2}{2\sqrt{\sqrt{(4k+3)(4k+1)}}} = \frac{1}{\sqrt{(4k+3)(4k+1)}}$

Therefore, each term in the summation is greater than or equal to 1, and since there are 2500 terms in the summation, the total sum is greater than or equal to 2500. This means that:

$\displaystyle \sum_{k=0}^{2499} \frac{1}{