Alternating summation and subtraction of square roots

Question

I encountered a problem, to find the integer part of: $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{3} + \sqrt{4}} +...+\frac{1}{\sqrt{99} + \sqrt{100}}$$.

I multiplied the conjugate of each denominator. Meaning, for $\frac{1}{\sqrt{a} + \sqrt{b}}$, I multiply $\sqrt{a} - \sqrt{b}$. I get $\sqrt{100} - \sqrt{99} + \sqrt{98} - \sqrt{97} + ... + \sqrt{2} - \sqrt{1}$. I am stuck here. How do I simplify this? Or is it that my method is wrong?

Thats what I did... you get $\sqrt{100} - \sqrt{99} + \sqrt{98} - \sqrt{97} + ... + \sqrt{2} - \sqrt{1}$ — QuIcKmAtHs, Jan 11 '18 at 08:42
Well I think the fact terms differing by one are paired close suggests telescoping.. — user3865391, Jan 11 '18 at 08:49
But I do not see how the terms can cancel out... @ClaudeLeibovici do you have any ideas? — QuIcKmAtHs, Jan 11 '18 at 08:51
I think $\sqrt{100} - \sqrt{99} + \sqrt{98} - \sqrt{97} + \cdots + \sqrt{2} - \sqrt{1}$ may be the answer. Its value $4.6323951\ldots$ does not suggest anything obvious to me — Henry, Jan 11 '18 at 08:54
If the largest term is $\sqrt{2n}$, so here $n=50$, then the sum is approximately $\sqrt{\frac{n}{2}}-0.38$ when $n$ is large — Henry, Jan 11 '18 at 09:04
Hmm perhaps can someone provide a solution that may be a close estimate? @Henry can you tell me how you got $4.6323951\ldots$? — QuIcKmAtHs, Jan 11 '18 at 09:49
@XcoderX in R: as expected n <- 1:50; sum(1/(sqrt(2*n-1)+sqrt(2*n))) gives the same result as n <- 1:50; sum(sqrt(2*n)-sqrt(2*n-1)) — Henry, Jan 11 '18 at 10:29
@Henry Your $-0.38$ can be identified as $\displaystyle(2\sqrt{2}-1),\zeta\left(-\frac12\right)$. — , Jan 11 '18 at 10:57

score 5 · Accepted Answer · 2018-01-11T11:04:29.477

5

For an approximation, we might start (using a not generally adopted, but not unfamiliar notation) with$$H^{(-1/2)}_n=1 + \sqrt{2} + \ldots + \sqrt{n}=\frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + \zeta\left(-\frac12\right)+o(1),$$ mentioned in this question ($\zeta$ is the famous Riemann zeta function).
Then, $$\sqrt{2n}-\sqrt{2n-1}\pm\ldots=-H^{(-1/2)}_{2n}+2\sqrt{2}\,H^{(-1/2)}_n=\frac12\sqrt{2n}+(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)+o(1).$$ For $n=50$, the main term would be $\displaystyle5+(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)=4.619895\ldots$, while the original sum is $4.6323951\ldots$, as @Henry pointed out, already, $\displaystyle(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)=-0.380105\ldots$ being slightly more accurate. The error (of the order $O(1/n)$) is no surprise.

edited Jan 11 '18 at 11:04

answered Jan 11 '18 at 10:46

Once again, somebody is happy that votes are anonymous, but how would that stop me laughing at you? The software of this site will stop you to make this a campaign. BTW, a downvote (fair or not) costs me just 2 points. – Jan 11 '18 at 10:53
1

To whoever downvoted: I think I know who you are, but let's pretend I don't. You are welcome to apply your own standards, but this post is very much ok. To the point that I am hard pressed to believe that the downvote is not personal. An ok answer to an ok question. What's there not to like? Bring in your sense of humor, step back and take a fresh look at a post. It is difficult to put past differences in the past, I have hard time doing that myself (which is why I won't do anything yet), but we should do our best. Both you and I. – Jyrki Lahtonen Jan 11 '18 at 11:27
@JyrkiLahtonen I agree this is a great answer. – QuIcKmAtHs Jan 11 '18 at 12:06
@XcoderX Don't you worry, the moderator doesn't suspect you. I'm glad you like my answer, though it doesn't help you, strictly speaking: I've just seen in the comments that you're supposed to find an approximation without a calculator. And most people wouldn't know how to compute $\zeta(-1/2)$ even with a calculator. So I'm working on estimates amenable to pencil-paper-work, but that will take some time. – Jan 11 '18 at 12:19
Haha, I did not think that. I just have that kind of feeling where you get downvoted for no apparent reason... Back to the question, estimates like root 2 can be easily made. – QuIcKmAtHs Jan 11 '18 at 12:21
Sorry I am unable to understand $(2\sqrt{2}-1),\zeta\left(-\frac12\right)=-0.380105\ldots$. How did you find $\zeta\left(-\frac12\right)$ ? Sorry again for such a basic question. – user8277998 Jan 11 '18 at 14:24
In my answer, I've linked a related question, since I didn't want to repeat what's written in more detail, there: https://math.stackexchange.com/questions/5676/how-closely-can-we-estimate-sum-i-0n-sqrti – Jan 11 '18 at 14:31
That doesn't answer my question of how did you find the value of $\zeta\left(-\frac12\right)$ ? – user8277998 Jan 11 '18 at 15:20
@user45914123 Sorry for the misunderstanding, I thought you meant the symbolic, not the numerical value. Well, that's easy, and free: http://www.wolframalpha.com/input/?i=zeta(-0.5) Normally, I'd use GP, but with Wolfram Alpha, it was quicker in that case. Of course, I could also transform it to the argument $3/2$ with the famous functional equation, and calculate that numerically with an alternating series. – Jan 11 '18 at 16:51
Can you show the other method ? Is it better than calculating all the square roots by hand ? – user8277998 Jan 11 '18 at 18:50
@user45914123 Sure I can, but that would be off topic, here. Ask the question, if you want to know, – Jan 11 '18 at 18:53
I can but I am afraid that I will get downvotes cause I am nothing of my work to show. By alternating series do you mean to calculate 4 or 5 terms of $\sum_{k=1}^\infty 1/k^{3/2}$; won't that be very difficult by hand ? – user8277998 Jan 11 '18 at 19:32
That's why I said "if you want to know". It's your decision. "Alternating series" means exactly that, an alternating series, you actually can express $\sum_{k=1}^\infty 1/k^{3/2}$ in terms of $\sum_{k=1}^\infty (-1)^{k-1}1/k^{3/2}$ And that's enough, moderators already complained about a private conversation, and they're right. If you have a question, ask it, but not in another person's question. – Jan 11 '18 at 21:50
@ProfessorVector Well it is your answer, you should improve it. The question is about finding the integer part without a calculator. So I asked how you approximated that fancy term, to which you said you used a calculator. Then I asked you to show the numerical method, that you won't do. I fail to see how this is an answer to the question asked. – user8277998 Jan 12 '18 at 21:42

score 2 · Answer 2 · answered Jan 11 '18 at 20:53

We want to find the integer part of $\sum_{x=1}^{50} f(x)$ where $f(x) = \sqrt{2x}-\sqrt{2x-1}$.

For decreasing $f(x)$, $$\int_a^{b+1} f(x) \; dx < \sum_{x=a}^b f(x) < \int_{a-1}^b f(x) \; dx$$ (It may help to make a sketch to see this.)

It's easier to work with the sum from $x=2$ to $50$ instead of $x=1$ to 50 because $f(x)$ is not defined when $x=0$. We can adjust for the missing $x=1$ term later.

By elementary calculus, $$\int f(x) \; dx = \frac{1}{3} (2x)^{3/2} - \frac{1}{3} (2x-1)^{3/2} + C$$ so $$\int_2^{51} f(x) \; dx < \sum_{x=2}^{50} f(x) < \int_{1}^{50} f(x) \; dx$$ yields $$4.10 < \sum_{x=2}^{50} f(x) < 4.38$$ Now to adjust for the missing $x=1$ term. Since $$\sum_{x=2}^{50} f(x) + \sqrt{2} - \sqrt{1} = \sum_{x=1}^{50} f(x)$$ we have $$4.10 + \sqrt{2} - \sqrt{1} < \sum_{x=1}^{50} f(x) < 4.38 + \sqrt{2} - \sqrt{1}$$ which yields $$4.51 < \sum_{x=1}^{50} f(x) < 4.79$$ so the integer part of the sum is $\boxed{4}$.

score 0 · Answer 3 · answered Jan 11 '18 at 11:33

0

$$\frac {1}{\sqrt n +\sqrt {n+1}} = \sqrt {n+1} - \sqrt {n}$$

Therefore the dsire sum is $ A-B$ where,

$$A= \sqrt 100+\sqrt 98 +....+\sqrt 2 = 338.047...$$ and $$B=\sqrt 99+\sqrt 97 +....+\sqrt 1 = 333.415....$$ Thus [A-B]=4

answered Jan 11 '18 at 11:33

Mohammad Riazi-Kermani

68,728

Ideally, that would be the answer, but unfortunately I cannot use a calculator. – QuIcKmAtHs Jan 11 '18 at 11:35

Alternating summation and subtraction of square roots

3 Answers3