# Show that the sum of the series is greater than 24

• Dec 14th 2012, 06:12 PM
Kaloda
Show that the sum of the series is greater than 24
Show that 1/(√1+√3)+1/(√5+√7)+1/(√9+√11) . . . + 1/(√9997+√9999) > 24.

• Dec 14th 2012, 10:00 PM
chiro
Re: Show that the sum of the series is greater than 24
Hey Kaloda.

Are you familiar with mathematical induction?
• Dec 14th 2012, 10:01 PM
abender
Re: Show that the sum of the series is greater than 24
Try taking advantage of conjugates. For example, the first term can be rewritten $\frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}} = \frac{\sqrt{3}-\sqrt{1}}{2}$.

In each case, the denominator is 2, so we can factor out a half.

The series eventually becomes $\frac{1}{2}\sum^{2499}_{k=0}[\sqrt{4k+3} - \sqrt{4k+1}]$.

Does this look nicer? Good luck.
-AB
• Dec 15th 2012, 02:26 AM
Kaloda
Re: Show that the sum of the series is greater than 24
@chiro:

No. Mathematical Induction is supposed to be our next topic.
• Dec 15th 2012, 06:56 AM
Kaloda
Re: Show that the sum of the series is greater than 24
Quote:

Originally Posted by abender
Try taking advantage of conjugates. For example, the first term can be rewritten $\frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}} = \frac{\sqrt{3}-\sqrt{1}}{2}$.

In each case, the denominator is 2, so we can factor out a half.

The series eventually becomes $\frac{1}{2}\sum^{2499}_{k=0}[\sqrt{4k+3} - \sqrt{4k+1}]$.

Does this look nicer? Good luck.
-AB

Any additional steps? I seem stucked.
• Dec 15th 2012, 01:15 PM
jakncoke
Re: Show that the sum of the series is greater than 24
I've been racking my head all day to do this without induction, i'm curious if anyone has an answer. I took a look at trying to do this with continued fractions but i had no idea what i was doing, if only i was ramanujan.
• Dec 15th 2012, 05:37 PM
Kaloda
Re: Show that the sum of the series is greater than 24
Can you show me the solution using Mathematical Induction?
I know the steps but I don't know how to apply it in cases that utilize summations.
• Dec 15th 2012, 07:26 PM
richard1234
Re: Show that the sum of the series is greater than 24
Quote:

Originally Posted by Kaloda
Any additional steps? I seem stucked.

Try writing out the first few terms of that sum. You'll obtain what is known as a telescoping sum.
• Dec 15th 2012, 07:48 PM
MarkFL
Re: Show that the sum of the series is greater than 24
If I were going to use induction, I would state the hypothesis:

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

or equivalently:

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

base case $P_0$:

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

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

Consider:

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

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

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

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

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

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

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

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

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

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

Now, adding this to the hypothesis, we have:

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

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

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

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

$\frac{1}{2}\sum_{k=0}^{2499}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{2500}-1=24$
• Dec 15th 2012, 09:17 PM
jakncoke
Re: Show that the sum of the series is greater than 24
I don't know why i didn't think of this earlier.

But if you have $\sum_{k=0}^{4998} \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}}$ then
take $f(k) = \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}}$
then $\int_0^{4999} f(k) dk \leq \sum_{k=0}^{4998} \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}} \leq \int_0^{4998} f(k) dk + f(0)$
So after computing the integral i got $\frac{1}{6}(-(2k+1)^{3/2} + 2k\sqrt(2k+3) + 3\sqrt(2k+3))$ , and after evaluating i got
so $49.99 \leq \sum_{k=0}^{4998} \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}} \leq 50.11$