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

AI Thread Summary
The discussion centers on proving the inequality $\sum_{k=0}^{2499} \frac{1}{\sqrt{4k+1}+\sqrt{4k+3}}>24$ without using induction. Participants suggest rationalizing the denominator and transforming the sum into a form involving square roots, leading to the expression $\frac{1}{2}\sum_{k=0}^{2499} \left(\sqrt{4k+3}-\sqrt{4k+1} \right)$. A proof by induction is outlined, establishing a hypothesis that connects the sums to a simpler form, ultimately demonstrating that the sum exceeds 24. Additionally, an alternative approach using geometric properties of square roots is mentioned, emphasizing the convexity of the function. The conversation highlights various methods to tackle the inequality, showcasing mathematical creativity and problem-solving techniques.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
MarkFL said:
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 $\frac12\Bigl(\sqrt{2500\tfrac14} - \sqrt{\tfrac14}\Bigr) > \frac12\bigl(50 - \tfrac12\bigr) = 24.75.$
 
MarkFL said:
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?
 
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top