# Thread: Is it true that...

1. If what you wrote is what I read, then yes, I am absolutely sure.

2. Originally Posted by wonderboy1953
Sorry I made the problem too hard:

$\displaystyle 1/(\sqrt{1} + \sqrt{2}) + 1/(\sqrt{2} + \sqrt{3}) + 1/(\sqrt{3} + \sqrt{4})+...+1/(\sqrt{498} + \sqrt{499}) + 1/(\sqrt{499} + \sqrt{500}) = 499?$

It's a great problem, algebra wise (that's a hint) and how would you go about proving that the left side of the equation equals the right side?
$\displaystyle \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =$

$\displaystyle = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} =$

$\displaystyle = \sum_{k=1}^n (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}-1$

Maybe this is what you meant?

3. Originally Posted by Unbeatable0
$\displaystyle \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =$

$\displaystyle = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} =$

$\displaystyle = \sum_{k=1}^n (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}-1$

Maybe this is what you meant?
Yes, that's it Unbeatable0. Part of the point to this problem is that it's really not as hard as it looks and rationalizing the denominator leads to cancellation of terms and the solution (I should have picked n = 528).

4. Originally Posted by Unbeatable0
$\displaystyle \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =$

$\displaystyle = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} =$

$\displaystyle = \sum_{k=1}^n (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}-1$

Maybe this is what you meant?
Yes, that's it Unbeatable0. Part of the point to this problem is that it's really not as hard as it looks and rationalizing the denominator leads to cancellation of terms and the solution (I should have picked n = 528).

5. Originally Posted by wonderboy1953
YES.

Methinks YOU should recheck.

6. Originally Posted by wonderboy1953
Sorry I made the problem too hard:

$\displaystyle 1/(\sqrt{1} + \sqrt{2}) + 1/(\sqrt{2} + \sqrt{3}) + 1/(\sqrt{3} + \sqrt{4})+...+1/(\sqrt{498} + \sqrt{499}) + 1/(\sqrt{499} + \sqrt{500}) = 499?$

It's a great problem, algebra wise (that's a hint) and how would you go about proving that the left side of the equation equals the right side?
LHS = 21.36067983

It's pretty clear that it's not 499.

Every term in the sum is less than 1 and there are 499 of them.

LHS = 21.36067983

It's pretty clear that it's not 499.

Every term in the sum is less than 1 and there are 499 of them.
Deadstar, I had set up the problem wrong (but it's a great problem).

8. Well.....what d'heck is the problem when set up RIGHT?
Why don't you post it again CORRECTLY, on a NEW thread, so we don't get all mixed up here!!

9. Originally Posted by Wilmer
Well.....what d'heck is the problem when set up RIGHT?
Why don't you post it again CORRECTLY, on a NEW thread, so we don't get all mixed up here!!
Two reasons: I don't want to repeat and I'm tired (at my library I have a time constraint which helps to wear me out).

10. Two reasons: I don't want to repeat and I'm tired (at my library I have a time constraint which helps to wear me out).
Repeat what ? You haven't even given the "great" problem yet.
Never mind, just post when you feel like it. And get it right

11. Originally Posted by wonderboy1953
Yes, that's it Unbeatable0. Part of the point to this problem is that it's really not as hard as it looks and rationalizing the denominator leads to cancellation of terms and the solution (I should have picked n = 528).
Dear Wonderboy1953,

Well I don't understand. If you want $\displaystyle \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =\sqrt{n+1}-1=499$

Then $\displaystyle \sqrt{n+1}=500\Rightarrow{n=249999}$

If you say that you should have picked n=528 then,

$\displaystyle \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =\sqrt{528+1}-1=22$

12. Originally Posted by Bacterius
Repeat what ? You haven't even given the "great" problem yet.
Never mind, just post when you feel like it. And get it right
(Forgive me moderators as I'm commanded to repeat and to get it right)

Is it true that:

$\displaystyle 1/(\sqrt1 + \sqrt2) + 1/(\sqrt2 + \sqrt3) + 1/(\sqrt3 + 1\sqrt4)+...+1/(\sqrt483 + \sqrt484) = 21?$

13. Originally Posted by wonderboy1953
(Forgive me moderators as I'm commanded to repeat and to get it right)
Is it true that:
$\displaystyle 1/(\sqrt1 + \sqrt2) + 1/(\sqrt2 + \sqrt3) + 1/(\sqrt3 + 1\sqrt4)+...+1/(\sqrt483 + \sqrt484) = 21?$
May I ask, Wboy, why does "posting it right" seem unimportant to you?