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

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

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

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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
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## 3 Answers

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

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

So:

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

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

Finally:

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

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

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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
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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 $$\frac{3}{4}$$

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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 $$s={1\over 2}\sum_{r=1}^\infty \Bigl({1\over r}-{1\over r+2}\Bigr)={1\over2}\bigl(1+{1\over2}\Bigr)={3\over4}\ .$$

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