Show that 1/(√1+√3)+1/(√5+√7)+1/(√9+√11) . . . + 1/(√9997+√9999) > 24.
I've no idea how to solve this. Please help. Ty.
Try taking advantage of conjugates. For example, the first term can be rewritten $\displaystyle \frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}} = \frac{\sqrt{3}-\sqrt{1}}{2}$.
In each case, the denominator is 2, so we can factor out a half.
The series eventually becomes $\displaystyle \frac{1}{2}\sum^{2499}_{k=0}[\sqrt{4k+3} - \sqrt{4k+1}]$.
Does this look nicer? Good luck.
-AB
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.
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:
$\displaystyle \frac{1}{2}\sum_{k=0}^{2499}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{2500}-1=24$
I don't know why i didn't think of this earlier.
But if you have $\displaystyle \sum_{k=0}^{4998} \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}} $ then
take $\displaystyle f(k) = \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}} $
then $\displaystyle \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 $\displaystyle \frac{1}{6}(-(2k+1)^{3/2} + 2k\sqrt(2k+3) + 3\sqrt(2k+3)) $ , and after evaluating i got
so $\displaystyle 49.99 \leq \sum_{k=0}^{4998} \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}} \leq 50.11 $