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

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MarkFL

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Feb 24, 2012
13,775
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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 $\frac12\Bigl(\sqrt{2500\tfrac14} - \sqrt{\tfrac14}\Bigr) > \frac12\bigl(50 - \tfrac12\bigr) = 24.75.$
 

Poly

Member
Nov 26, 2012
32
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?
 
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MarkFL

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Feb 24, 2012
13,775
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.