# Alternating summation and subtraction of square roots

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 8:42
• And it is a - b @Wauzl – QuIcKmAtHs Jan 11 '18 at 8:43
• But I do not see how the terms can cancel out... @ClaudeLeibovici do you have any ideas? – QuIcKmAtHs Jan 11 '18 at 8:51
• 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 8:54
• @Henry Your $-0.38$ can be identified as $\displaystyle(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)$. – Professor Vector Jan 11 '18 at 10:57

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.

• 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. – Professor Vector Jan 11 '18 at 10:53
• 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. – Professor Vector 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

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 $\boxed{4}$.

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

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

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