Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to prove the existence and the exact value of the following series:

$$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$$

share|cite|improve this question
Hint: When is $\lfloor\sqrt{n+1}\rfloor \neq \lfloor \sqrt{n}\rfloor$? – Thomas Andrews Feb 22 '12 at 14:58
Think about the first quite a few terms. It is easy to write them down explicitly. You should get something familiar, a famous series. – André Nicolas Feb 22 '12 at 15:06
Thank you for your help. – Chon Feb 22 '12 at 16:19
@Chon I'm guessing by exact you mean closed form, right? – user285523 Nov 11 '15 at 4:35
up vote 10 down vote accepted

If $ n+1 $ is a square, $\lfloor \sqrt{n+1} \rfloor=\sqrt{n+1} $

$$ 0<\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<1$$

So: $$ \sqrt{n+1}-1<\sqrt{n}<\sqrt{n+1} $$

So: $$ \lfloor \sqrt{n} \rfloor= \sqrt{n+1}-1$$

$$ \lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor=1 $$

I have proved:

$n+1$ is a square $\Longrightarrow \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor$

Now I must show that

$ \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor \Longrightarrow $ $n+1$ is a square

If $ n+1$ is not a square:

If $ n $ is a square, $\lfloor \sqrt{n} \rfloor=\sqrt{n}$. As $$ \sqrt{n+1}-1<\sqrt{n}<\sqrt{n+1} $$

$ \lfloor \sqrt{n+1} \rfloor= \sqrt{n} $

So: $ \lfloor \sqrt{n+1} \rfloor= \lfloor \sqrt{n} \rfloor $

If $ n $ is not a square, there exists $a\in \mathbb{N} $ such that

$$ a^2<n<n+1<(a+1)^2$$


$$ a<\sqrt{n}<\sqrt{n+1}<a+1$$

So : $$ \lfloor \sqrt{n+1} \rfloor= \lfloor \sqrt{n} \rfloor $$


$n+1$ is a square $\Longleftrightarrow \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor$

So: $$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}=\sum_{n=2}^{\infty} \frac{1}{n^2-1}=\cdots=\frac{3}{4}$$

share|cite|improve this answer
Good. You could have saved some space. For example, if $n+1$ is not a square, then there exists an integer $a$ such that $a^2\le n<n+1<(a+1)^2$. So you don't need to consider $n$ a square, not a square separately. – André Nicolas Feb 22 '12 at 16:59
You are right, thank you! – Chon Feb 22 '12 at 17:14
Your last sum should start at $2$, not $1$. – David Mitra Feb 22 '12 at 17:26

Work on Thomas Andrews's hint (That is a very good hint)

Let me give a following hint

$$ \left\lfloor \sqrt{(3+1)} \right\rfloor = 2, \hspace{3pt} \left\lfloor \sqrt{(2+1)} \right\rfloor = 1, \text{ why?} $$

$$ \left\lfloor \sqrt{(8+1)} \right\rfloor = 3, \hspace{3pt} \left\lfloor \sqrt{(7+1)} \right\rfloor = 2, \text{ why?} $$

Try working towards these observed values, and in general for what values are they not equal?

The answer is $$\frac{3}{4}$$

share|cite|improve this answer

As all terms of the given series $s=\sum_{n=1}^\infty a_n$ are $\geq0$ we may collect them in packets and write $$s=\sum_{r=1}^\infty\left(\sum\nolimits_{r^2\leq n<(r+1)^2} a_n\right)\ .$$ Note that in the inner sum only the last term, corresponding to $n=(r+1)^2-1$, is nonzero and has the value $${1\over(r+1)^2-1}={1\over (r+2) r}={1\over2}\Bigl({1\over r}-{1\over r+2}\Bigr)\ .$$ It follows that the outer sum is a telescoping series, and we obtain $$s={1\over 2}\sum_{r=1}^\infty \Bigl({1\over r}-{1\over r+2}\Bigr)={1\over2}\bigl(1+{1\over2}\Bigr)={3\over4}\ .$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.