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

Question

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

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

Answer 1

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

Answer 2

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

Answer 3

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